DC Circuits - Problem Solving
Example problem on ohm's law: the basic circuit.
An emf source of \(6.0V\) is connected to a purely resistive lamp and a current of \(2.0\) amperes flows. All the wires are resistance-free. What is the resistance of the lamp?
- Where in the circuit does the gain in potential energy occur?
- Where in the circuit does the loss of potential energy occur?
- What is Ohm's Law?
The gain of potential energy occurs as a charge passes through the battery, that is, it gains a potential of \(\varepsilon = 6.0V\) . No energy is lost to the wires, since they are assumed to be resistance-free. By conservation of energy, the potential that was gained (i.e. \(\varepsilon= V = 6.0V\) ) must be lost in the resistor. So, by Ohm's Law:
\(V = I R\)
\(R = 3.0 \Omega\)
Example of Problem on Resistors in Series
The current flowing in a circuit containing four resistors connected in series is \(I = 1.0 A\) . The potential drops across the first, second and third resistors are, respectively: \(V = 5 V\) , \(V = 8 V\) and \(V = 7 V\) . The equivalent resistance of the circuit is \(R = 30 \Omega\) .
Find the total voltage supplied by the battery, and also current, voltage drop, and resistance of each resistor in the circuit.
- How are resistors related when connected in series?
- What is true about potential drops of resistors when connected in series?
- You will need to use Ohm's Law.
First, let's label the diagram with the information given in the question.
There are several ways of solving this problem (see alternate solutions ), but this tutorial will only go through one of these ways.
Because the resistors are connected in series, then the same current flows through each one. Using the Ohm's Law, we can find the resistances of the first, second and third resistors.
\(R_1 = \frac {V_1}{I}, R_2 = \frac{V_2}{I} , R_3 = \frac {V_3}{I}\)
\(R_1 = \frac {5.0}{1.0} = 5.0 \quad \Omega R_2 = \frac{8.0}{1.0} = 8.0 \quad \Omega R_3 = \frac {7.0}{1.0} = 7.0\Omega\)
Now, using the equivalent resistance, we can find the resistance in the fourth resistor. This is a series circuit, so the equivalent resistance is the sum of the individual resistances.
\(R_ \mathrm {equivalent} = R_1 + R_2 +R_3+R_4\)
\(R_4 = R _\mathrm {equivalent} - ( R_1+R_2+R_3)\)
\(R_4 = 30 - (5.0 + 8.0 + 7.0) = 10 \Omega\)
The current flowing through the fourth resistor is also \(I=1.0A\) . Using Ohm's Law again, we find the voltage across this resistor.
\(V_4 = I \cdot R_4\)
\(V_4 = (1.0)\cdot (10) = 10V\)
The total voltage supplied by the battery must equal to the total voltage drop across the circuit (this is known as Kirchhoff's Voltage Law ). So, we must sum up the voltage drops across the resistors.
\(V = V_1 +V_2 +V_3 +V_4\)
\(V = 5.0+8.0+7.0+10 = 30 V\)
Example Problem on Resistors in Parallel
In the following schematic diagram, find the total current, I.
- You will need Ohm's Law.
- How are resistors related when connected in parallel?
- What is the potential drop across each resistor?
- How does current behave in parallel branches?
We know the total potential of this circuit,
\(\varepsilon = 12.0 V\)
So, between points \(A\) and \(B\) , the potential must drop \(12.0V\) . Also, the potential drop across branches of a circuit are equal. That is,
\(V_1 = V_2 = V_3 = \varepsilon = 12.0V\)
We can use Ohm's Law
\(V = IR\) or \(I = V/R\)
to find the current across each resistor.
\(I_1 = \frac {V_1}{R_1} = \frac {12.0V}{2.0 \Omega} = 6.0A\)
\(I_2 = \frac{V_2}{R_2} = \frac{12.0V}{3.0 \Omega} = 4.0A\)
\(I_3 = \frac {V_3}{R_3} = \frac {12.0V}{6.0 \Omega} = 2.0A\)
Recall that the currents through branches of a parallel circuit add to give the total current. That is, the total current 'splits up' so that part of the total current travels down each branch. Because of conservation of charge, the sum of the currents in each branch must equal the amount going into the branch. (This is Kirchhoff's Current Law.)
So, adding up the three currents, we get:
\(I = I_1 +I_2 + I_3\)
\(= 6.0 + 4.0 +2.0 = 12.0 A\)
So, the total current is \( I = 12.0A\) .
Example Problems on Resistors in Combination Circuits
- Which resistors are in parallel and which are in series?
- Is this circuit composed of small groups of parallel resistors, all connected in series? Or is it composed of groups of series resistors, connected in parallel?
This circuit is composed of 3 'elements' connected in series: the group of parallel resistors between \(A\) and \(B\) , the single resistor \(R_3\) , and the group of parallel resistors between \(C\) and \(D\) .
First, we will find the equivalent resistance between \(A\) and \(B\) .
Here, we have two resistors, \(R_1\) and \(R_2\) , connected in parallel. Using the formula for resistors connected in parallel:
\(\frac {1}{R_\mathrm{equivalent}} = \sum \frac{1}{R_i}\)
we can find the equivalent resistance between points \(A\) and \(B\) . Let's call this equivalent resistance \(R_{AB}\) .
\(\frac {1}{R_{AB}} = \frac {1}{R_1} + \frac {1}{R_2}\)
\(= \frac {1}{10.0}+ \frac{1}{4.0}\)
\(R_{AB} = 2.857 \Omega\)
Now, we'll find the equivalent resistance between \(C\) and \(D\) , and will call it \(R_{CD}\) . Using the equation from above for resistors connected in parallel,
\(\frac {1}{R_{CD}} = \frac {1}{R_4} + \frac {1}{R_5}\)
\(\frac {1}{R_{CD}} = \frac{1}{8.0}+\frac{1}{1.0}\)
\(\frac {1}{R_{CD}} = 1.125\)
\(R_{CD} = 0.889 \Omega\)
Replacing the two parallel sections with their equivalent resistances, and redrawing the circuit, we get the circuit in Figure 2. We see that there are three resistances connected in series: \(R_{AB}\) , \(R_3\) , and \(R_{CD}\) . Using the formula for resistors in series,
we can find the equivalent resistance of the circuit.
So the equivalent resistance of this circuit is \(R = 6.7 \Omega\) .
Figure 7 shows part of a circuit. It consists of resistors combined in both parallel and series configurations. Find the equivalent resistance.
- What is the equivalent resistance for resistors in parallel?
In this partial circuit, there are three main branches, branch \(AB\) , branch \(CD\) , branch \(EF\) . As you can see, branch \(AB\) contains two resistors in series, \(R_1\) and \(R_2\) . Branch \(CD\) has just one resistor, \(R_3\) . Finally, there are two resistors in branch \(EF\) .
Let's look at branch \(AB\) first. We will simplify this branch by finding the equivalent resistance between \(A\) and \(B\) . Note that \(R_1\) is connected in series with \(R_2\) . Using the equation for resistors in series
\(R_\mathrm {equivalent} = \sum R_i\)
we can find \(R_{AB}\) .
\(R_{AB} = R_1 + R_2\)
\(R_{AB} = 1.0 + 2.0\)
\(R_{AB} = 3.0 \Omega\)
Now, in branch \(CD\) there is only one resistor, so this branch cannot be simplified further.
In branch \(EF\) , however, there are two resistors, connected in series with one another. Using the equation for resistors in series, we can find the equivalent resistance in branch \(EF\) , \(R_{EF}\) .
\(R_{EF} = R_4 + R_5\)
\(R_{EF} = 4.0 + 5.0\)
\(R_{EF} = 9.0 \)
We can redraw Circuit 2 using \(R_{AB}\) , \(R_3\) , and \(R_{EF}\) , as seen in Figure 8. This circuit has been simplified to a parallel circuit, with three resistances in parallel. Using the formula for resistors connected in parallel
\(\frac {1}{R_\mathrm {equivalent}} = \sum \frac {1}{R_i}\)
we can find the equivalent resistance of these branches.
\(\frac {1}{R} = \frac {1}{R_{AB}} + \frac {1}{R_3}+ \frac{1}{R_{EF}}\)
\(\frac{1}{R} = \frac{1}{3.0} + \frac{1}{3.0}+ \frac{1}{9.0}\)
\(R = 1.286 \Omega\)
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AP Physics 2: Circuits Practice Problems with Answers
Here, a few sample AP Physics C problems on circuits are gathered and solved. These questions involve the various configuration of capacitors, resistances, and electric power in a DC circuit.
Circuit Practice Problems
Problem (1): In the circuit below, find the equivalent resistance between points $A$ and $B$.
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Recall that the equivalent of two resistors in series is $R_{eq}=R_1+R_2$ and in parallel is $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$.
The two resistors $3-{\rm \Omega}$ are in series, so their equivalent resistance is \[R_{3,3}=R_3+R_3=3+3=6\,{\rm \Omega}\] It's obvious that resistors of $12\,{\rm \Omega}$ and $6\,{\rm \Omega}$ are in parallel, so their effective or equivalent resistance is \begin{align*}\frac{1}{R_{12,6}}&=\frac{1}{R_{12}}+\frac{1}{R_6}\\\\&=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}\\\\\Rightarrow R_{12,6}&=4\quad{\rm \Omega}\end{align*} Now, $R_{12,6}$ is in series with the resistor of $8\,{\rm \Omega}$ which both have a effective resistance of \[R_{12,6,8}=R_{12,6}+R_8=4+8=12\,{\rm \Omega}\] In the end, we are left two resistance of $12\,{\rm \Omega}$ and $6\,{\rm \Omega}$ which are placed in parallel in the circuit (Because each are positioned on one side of points $A$ and $B$). Thus, the overall resistance of the circuit is determined as below \begin{align*}\frac{1}{R_{eq}}&=\frac{1}{R_{3,3}}+\frac{1}{R_{12,6,8}}\\\\&=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}\\\\\Rightarrow R_{eq}&=4\quad{\rm \Omega}\end{align*} Note: When two resistors $R_1$ and $R_2$ are in parallel, we can use the simple formula below to avoid using reciprocals \[R_{eq}=\frac{R_1 R_2}{R_1+R_2}\]
Solution : The two resistances of $120\,{\rm \Omega}$ and $40\,{\rm \Omega}$ are in parallel, so we can remove these two resistors and replace them with their equivalent resistance which is \[R_{120,40}=\frac{R_1 R_2}{R_1+R_2}=\frac{120\times 40}{120+40}=30\,{\rm \Omega}\] Two resistors $5\,{\rm \Omega}$ and $R$ in the leftmost vertical branch are in series, so their equivalent resistance is $R_{R,5}=R+5$. These two latter equivalent resistances are also in parallel having an effective resistance of \[\frac{30\times (R+5)}{30+(R+5)}=\frac{30R+150}{R+35}\] In the end, this resistance is placed in series with the $R$ resistor. Thus, the overall resistance of the circuit is obtained as \[R_{eq}=R+\frac{30\times (R+5)}{30+(R+5)}\] Setting this $40\,{\rm \Omega}$ and Solving for $R$, we get the value of the unknown resistor \begin{gather*} R+\frac{30R+150}{R+35}=40 \\\\ \frac{R(R+35)+30R+150}{R+35}=40 \\\\ \Rightarrow 40R+1400=R^2+35R+30R+150 \\\\ \Rightarrow R^2+25R-1250=0 \end{gather*} To solve the quadratic equation $ax^2+bx^2+c=0$ where $a,b,c$ are some constants, use the following equation \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] In this case, $a=1,b=25,c=-1250$, substituting these numerical values into the above equation, we obtain two solutions as \[R=\frac{2(1)\pm\sqrt{25^2-4(1)(-1250)}}{2(1)}\] The plus sign gives $\boxed{R=25\,{\rm \Omega}}$ and the minus sign gives $R=-50\,{\rm \Omega}$ which is incorrect (since the resistance can not be a negative value).
Problem (3): In the following circuit, how much must $R_3$ be so that the equivalent resistance between $A$ and $B$ is $R_1$?
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Problem (4): In the circuit below, how much current passes through the $1.5-{\rm \Omega}$ resistor?
Solution : In this kind of question, first of all, find the total equivalent resistance of the circuit, then using Ohm's law find the amount of total current delivered by the battery.
Two resistors $3\,{\rm \Omega}$ and $1.5\,{\rm \Omega}$ are in parallel, so their equivalent resistance is \[R_{eq}=\frac{3\times 1.5}{3+1.5}=1\,{\rm \Omega}\] This effective resistor is placed in series with the remaining external resistance of the circuit, $3.5\,{\rm \Omega}$. With this simplification, the circuit involves the resistances of $1\,{\rm \Omega}$, $3.5\,{\rm \Omega}$, and the internal resistance of the battery. Thus, combine them to find the total equivalent resistance of the circuit \[R_t=1+3.5=4.5\,{\rm \Omega}\]
When a circuit is connected to a source of emf (battery), the total current through the battery is found by the following formula \[I_t=\frac{\mathcal{E}}{r+R_{eq}}=\frac{6}{1.5+4.5}=1\,{\rm A}\] Thus, through each resistor a current of $1\,{\rm A}$ flows. The next step is to find the current through the $1.5\,{\rm \Omega}$ resistor which is part of $1\,{\rm \Omega}$ combined resistor.
The two $1.5\,{\rm \Omega}$ and $3\,{\rm \Omega}$ resistors are in parallel with each other. Recall that the voltages across resistors in parallel are the same i.e. $V_3=V_{1.5}$. Using this fact and Ohm's law, we find a relation between the current through these two resistors as below \begin{gather*} V_3=V_{1.5} \\\\ I_3 R_3=I_{1.5}R_{1.5}\\\\ 3I_3=1.5I_{1.5} \\\\ \Rightarrow \boxed{I_3=\frac 12 I_{1.5}}\end{gather*} The $1-{\rm A}$ total current provided by the battery when reaches the junction $a$, splits into the two currents $I_3$ and $I_{1.5}$. So, their sum must be the same total current (i.e., $I_t=I_3+I_{1.5}$).
In this case, we have $I_{1.5}+I_3=1\,{\rm A}$. Therefore, substituting the boxed relation above into this, we get \begin{gather*} I_{1.5}+\frac 12 I_{1.5}=1 \\\\ \Rightarrow (1.5)I_{1.5}=1 \\\\ \Rightarrow I_{1.5}=\frac{1}{1.5}=\frac 23 \,{\rm A}\end{gather*} Thus, the current through $1.5\,{\rm \Omega}$ is $1.5\,{\rm A}$.
Further reading: Ohm's law Practice problems with solutions
The power dissipated by resistance is found by the following formulas \[P=R i^2=\frac{V^2}{R}=Vi\] where $V$ is the voltage across the resistance and $i$ is the current through it.
In this circuit the $33-{\rm \Omega}$ resistor is connected directly to the battery, so all the current delivered by the battery flows through it. In other words, the current passing through this resistor gives the total current of the circuit.
Using the power formula $P=R i^2$, and solving for $i$, we can find the current flowing from the $33-{\rm \Omega}$ resistor \[i=\sqrt{\frac{P}{R}}=\sqrt{\frac{0.8}{33}}=0.156\,{\rm A}\] This is the total current flowing from the battery $I_t=0.156\,{\rm A}$. Now we must find the equivalent resistance of the circuit.
The two resistors of $68\,{\rm \Omega}$ and $85\,{\rm \Omega}$ are in parallel, so their equivalent resistance is found as below \[R_{eq}=\frac{68\times 85}{68+85}=37.7\,{\rm \Omega}\] This new resistor is placed in series with the $33-{\rm \Omega}$ resistor, so the total equivalent resistance of the circuit is \[R_t=33+37.7=70.7\quad {\rm \Omega}\] Given the total current and resistance of the circuit, one can find the voltage across the battery using Ohm's law as below \[V_t=I_t R_t=0.15\times 70.7=11\,{\rm V}\]
The resistors $2-{\rm \Omega}$ and $6-{\rm \Omega}$ are in series, so the currents through them are the same. Given the power dissipated by $2-{\rm \Omega}$ resistor and using the power formula $P=Ri^2$, we can find the current flowing from this resistor \[i=\sqrt{\frac{P}{R}}=\sqrt{\frac{18}{2}}=3\,{\rm A}\] Thus, $I_2=I_4=3\,{\rm A}$. The equivalent resistance of these two is \[R_{2,6}=R_2+R_6=2+6=8\,{\rm \Omega}\] The total current and equivalent resistance of the upper branch are known, so using Ohm's law, we can find the voltage across this branch as \[V_{up}=IR=3\times 8=24\,{\rm V}\] The lower branch also has the same voltage as the up branch according to this rule that the voltages across parallel branches are the same. Therefore, \[V_{upper}=V_{lower}=V_{ef}=24\,{\rm V}\]
Further Reading: Kirchhoff's law solved problems for the AP Physics C exam
Problem (7): In the circuit shown below, the voltage drop across the internal resistance and the $2.5-{\rm \Omega}$ resistor are $0.25\,{\rm V}$ and $1.25\,{\rm V}$, respectively. Find the emf and internal resistance of the battery.
Solution : The potential drop across the internal resistance of a battery is defined as $V_r=Ir$, and across the resistor external to the battery is $V_R=IR$ where $I$ is the total current through the circuit.
Using these two facts and their given numerical values, we can find their ratio \begin{align*} \frac{V_r}{V_R}&=\frac{Ir}{IR}\\\\ \frac{0.25}{1.25}&=\frac{r}{2.5}\\\\ \Rightarrow r&=0.5\,{\rm \Omega}\end{align*} Thus, the internal resistance of the battery is $\boxed{r=0.5\,{\rm \Omega}}$.
Problem (8): In the circuit below, when the switch is opened, the voltmeter shows $12\,{\rm V}$, and when it is closed, the voltmeter reading is $10\,{\rm V}$. What are the emf and the internal resistance of the battery?
Solution : Remember that in a circuit including a source of emf with an internal resistance of $r$ and an equivalent resistance of $R_{eq}$ due to other resistances external to the battery, the total current flowing through the circuit is found by the formula below \[I=\frac{\mathcal{E}}{r+R_{eq}}\] Here, the voltmeter is connected across the terminals of the battery, so it shows the battery's terminal voltage. Always, the terminal voltage is given by $\Delta V=\mathcal{E}-Ir$, where $I$ is the total current through the circuit.
When the switch is opened, the middle branch is removed from the circuit, so the equivalent resistance of the circuit equals $R_{eq}=R$. Correspondingly, the current in this case is \[I_O=\frac{\mathcal{E}}{r+R_{eq}}=\frac{\mathcal{E}}{r+R}\] where $I_O$ is the current when the switch is open. By opening the switch, the voltmeter reading is also found as \begin{align*} V_O&=\mathcal{E}-I_O r\\\\ &=\mathcal{E}-\left(\frac{\mathcal{E}}{r+R}\right)r \\\\&=\frac{\mathcal{E}R}{R+r}\end{align*} Now we turn to the case of closing the switch $S$. In this case, the equivalent resistance is \[R_{eq}=\frac{R\times R}{R+R}=\frac R2\] Since the resistors are in parallel. The ammeter shows the total current flowing from the battery whose value is \[I_C=\frac{\mathcal{E}}{r+R_{eq}}=\frac{\mathcal{E}}{r+R/2}\] On the other side, as mentioned, the voltmeter indicates the voltage across the battery or terminal voltage with the following value \begin{align*} V_C&=\mathcal{E}-I_C r\\\\ &=\mathcal{E}-\left(\frac{\mathcal{E}}{r+R/2}\right)r \\\\&=\frac{\mathcal{E}R}{R+2r}\end{align*} Now we compare the currents and voltages before and after closing the switch $K$: \begin{gather*} \frac{\mathcal{E}}{r+R/2}> \frac{\mathcal{E}}{r+R} \\\\ \Rightarrow \boxed{I_C > I_O} \\\\ \frac{\mathcal{E}R}{R+r} > \frac{\mathcal{E}R}{R+2r} \\\\\Rightarrow \boxed{V_O > V_C} \end{gather*}
Problem (10): (a) Four $3.2-{\rm \mu F}$ capacitors are connected in series. What is the equivalent capacitance? (b) Assuming now they are connected in parallel, what is their equivalent capacitance?
Problem (11): In the following circuit, what is the potential difference across the capacitor? a. 8V b. 10V c. 6V d. 15V
Solution : As can be seen from the circuit, the two $4-{\rm \Omega}$ resistors are in parallel, so we compute their combined resistance and replace them with those. \[R_{eq}=\frac{R_1 R_2}{R_1+R_2}=\frac{4\times 4}{4+4}=2\,{\rm \Omega}\] Note that when a fully charged capacitor is present in an electric circuit, that branch acts as an open circuit, which means the current no longer flows through it. Because, in practice, the capacitor behaves like a resistor with infinite resistance.
On the other side, the current delivered by a source of emf (battery) with internal resistance $r$ and external resistance $R_{eq}$ is found by \[I=\frac{\mathcal{E}}{r+R_{eq}}=\frac{12}{1+2}=4\,{\rm A}\] All this current flows through the equivalent resistance of $2\,{\rm \Omega}$. Given that, we can find the potential drop across this combined resistor, using Ohm's law, as \[V=IR=4\times 2=8\,{\rm V}\] Now pay attention to this fact that the capacitor and $2-{\rm \Omega}$ equivalent resistance are in parallel which means the voltage across the capacitor is the same as the voltage across the resistor (i.e., $V_C=V_{2R}$).
Problem (12): In the circuit diagram shown below, how much charge is on the plates of the capacitor stored?
Solution : The charge stored on the plates of a capacitor is related to the voltage across it by $Q=C\Delta V$. The capacitance of the capacitor is known, the only thing that we must find is the potential drop (voltage) across it.
On the other side, you can see that the capacitor is connected in parallel with the $4-{\rm \Omega}$ resistor, which means that the voltage difference across the capacitor is the same as the resistor. So, our task is to find the voltage across the $4-{\rm \Omega}$ resistor, $\Delta V_4=IR$, which is subsequently needed to find the current through that resistor.
As a rule of thumb, ignore that branch of the circuit that includes the capacitor since the capacitors in an electric circuit (when it is fully charged) act as a resistor with an infinite resistor and do open the circuit.
With this long introduction, let's move on to solving the problem.
By ignoring the capacitor, the three $4-{\rm \Omega}$ resistors are in series with a equivalent resistance of $R_{eq}=3(4)=12\,{\rm \Omega}$. This new resistor is in parallel with the other $4-{\rm \Omega}$ resistor in the middle branch.
Problem (13): In the circuit shown below, the capacitor does not have any charge on its plates. Immediately after closing the switch $S$, what are the currents $I_1$ and $I_2$, respectively? a. 1, 1 b. 1.5, 0 c. 0, 1.5 d. 1.5, 3
Conversely, after passing a long time, it behaves like a resistor with infinite resistance.
Initially, the capacitor has no charge, so the potential difference across it, using the capacitance formula $\Delta V=\frac{Q}{C}$, is zero. On the other side, this capacitor is in parallel with the $5-{\rm \Omega}$ resistor.
Recall that when two electronic components (e.g., a capacitor or resistor) are in parallel, the potential drop across them is the same. According to this rule, the potential difference across the $5-{\rm \Omega}$ resistor is also zero. Applying Ohm's law, $\Delta V=IR$, we find that the current flowing through this resistor must also be zero.
Consequently, the current $I_2$ shown in the circuit, immediately after closing the switch, must be zero. In this situation, the $5-{\rm \Omega}$ resistor is removed and the circuit converts to the following
Note that in all AP Physics circuit multiple-choice problems, just after placing an initially uncharged capacitor in a circuit (like immediately after closing a switch in the circuit), it acts as a typical wire with zero resistance, and subsequently, all branches in parallel with it can be removed from the circuit.
Problem (14): In the circuit below, the capacitor is initially uncharged. A long time after closing the switch $S$, find the currents $I_1$ and $I_2$. a. 1, 1 b. 1.5, 0 c. 0, 1.5 d. 1.5, 3
After a long time, when the switch is closed, the charge accumulation on the capacitor reaches its maximum value. At this moment, this charge, which is built on the plates of the capacitor, impedes completely the charge flowing by the battery (precisely, the current through this branch).
So, the current flowing through this branch is stopped and all total current delivered by the battery (i.e., $I_1$) reaching the junction $A$, flows into the branch where the $5-{\rm \Omega}$ resistor is presented. In other words, $I_1=I_2$.
On the other side, the current flowing through the circuit is found as \[I_t=\frac{\mathcal{E}}{r+R_{eq}}=\frac{15}{0+15}=1\,{\rm A}\] Therefore, \[I_1=I_2=I_t=1\,{\rm A}\] The correct answer is a.
In all AP Physics circuit multiple-choice questions, note that, after a long time of placing an initially uncharged capacitor in a circuit, it acts as an open circuit, and we can remove that branch from the circuit.
Author : Dr. Ali Nemati Page Published : 10/13/2021
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Physics Problems with Solutions
Dc circuits calculators of currents and voltages, 1 - voltages and currents formulas in circuit 1, voltages and currents calculator for circuit 2, 2 - voltages and currents formulas in circuit 2, 3 - voltages and currents formulas in circuit 3, voltages and currents calculator for circuit 3, more references and links, popular pages.
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Unit: Circuit analysis
Circuit elements.
- Ideal circuit elements (Opens a modal)
- Ideal sources (Opens a modal)
- Ideal elements and sources (Opens a modal)
- Real-world circuit elements (Opens a modal)
- Circuit terminology (Opens a modal)
- Sign convention for passive components (Opens a modal)
- Sign convention for passive components and sources (Opens a modal)
Resistor circuits
- Series resistors (Opens a modal)
- Parallel resistors (part 1) (Opens a modal)
- Parallel resistors (part 2) (Opens a modal)
- Parallel resistors (part 3) (Opens a modal)
- Parallel resistors (Opens a modal)
- Parallel conductance (Opens a modal)
- Simplifying resistor networks (Opens a modal)
- Delta-Wye resistor networks (Opens a modal)
- Voltage divider (Opens a modal)
- Analyzing a resistor circuit with two batteries (Opens a modal)
- Series and parallel resistors 8 questions Practice
DC circuit analysis
- Circuit analysis overview (Opens a modal)
- Kirchhoff's current law (Opens a modal)
- Kirchhoff's voltage law (Opens a modal)
- Kirchhoff's laws (Opens a modal)
- Labeling voltages (Opens a modal)
- Application of the fundamental laws (setup) (Opens a modal)
- Application of the fundamental laws (solve) (Opens a modal)
- Application of the fundamental laws (Opens a modal)
- Node voltage method (steps 1 to 4) (Opens a modal)
- Node voltage method (step 5) (Opens a modal)
- Node voltage method (Opens a modal)
- Mesh current method (steps 1 to 3) (Opens a modal)
- Mesh current method (step 4) (Opens a modal)
- Mesh current method (Opens a modal)
- Loop current method (Opens a modal)
- Number of required equations (Opens a modal)
- Linearity (Opens a modal)
- Superposition (Opens a modal)
Natural and forced response
- Capacitor i-v equations (Opens a modal)
- A capacitor integrates current (Opens a modal)
- Capacitor i-v equation in action (Opens a modal)
- Inductor equations (Opens a modal)
- Inductor kickback (1 of 2) (Opens a modal)
- Inductor kickback (2 of 2) (Opens a modal)
- Inductor i-v equation in action (Opens a modal)
- RC natural response - intuition (Opens a modal)
- RC natural response - derivation (Opens a modal)
- RC natural response - example (Opens a modal)
- RC natural response (Opens a modal)
- RC step response - intuition (Opens a modal)
- RC step response setup (1 of 3) (Opens a modal)
- RC step response solve (2 of 3) (Opens a modal)
- RC step response example (3 of 3) (Opens a modal)
- RC step response (Opens a modal)
- RL natural response (Opens a modal)
- Sketching exponentials (Opens a modal)
- Sketching exponentials - examples (Opens a modal)
- LC natural response intuition 1 (Opens a modal)
- LC natural response intuition 2 (Opens a modal)
- LC natural response derivation 1 (Opens a modal)
- LC natural response derivation 2 (Opens a modal)
- LC natural response derivation 3 (Opens a modal)
- LC natural response derivation 4 (Opens a modal)
- LC natural response example (Opens a modal)
- LC natural response (Opens a modal)
- LC natural response - derivation (Opens a modal)
- RLC natural response - intuition (Opens a modal)
- RLC natural response - derivation (Opens a modal)
- RLC natural response - variations (Opens a modal)
AC circuit analysis
- AC analysis intro 1 (Opens a modal)
- AC analysis intro 2 (Opens a modal)
- Trigonometry review (Opens a modal)
- Sine and cosine come from circles (Opens a modal)
- Sine of time (Opens a modal)
- Sine and cosine from rotating vector (Opens a modal)
- Lead Lag (Opens a modal)
- Complex numbers (Opens a modal)
- Multiplying by j is rotation (Opens a modal)
- Complex rotation (Opens a modal)
- Euler's formula (Opens a modal)
- Complex exponential magnitude (Opens a modal)
- Complex exponentials spin (Opens a modal)
- Euler's sine wave (Opens a modal)
- Euler's cosine wave (Opens a modal)
- Negative frequency (Opens a modal)
- AC analysis superposition (Opens a modal)
- Impedance (Opens a modal)
- Impedance vs frequency (Opens a modal)
- ELI the ICE man (Opens a modal)
- Impedance of simple networks (Opens a modal)
- KVL in the frequency domain (Opens a modal)
About this unit
Diode Circuit Analysis
Published Mar 03, 2020
So we've learned about diodes in previous tutorials. But today we're going to be solving circuits with diodes and then we're going to be finding what the quiescent or Q-point is for those diodes and we're going to be talking about the benefits of each of these methods.
The Q-point is basically that point around which you don't expect a whole lot of variation in the current or the voltage. And with diodes, there's a couple of different ways to approach it. And we're going to go over them really quick, we're going to talk about two of them, talk about why we're not going to be doing them. And then we're going to do two examples with two of the other methods. So again, we're looking for the voltage across it and the current through it.
Load Line Analysis
So the first thing that you can do, and one of the ones that we don't recommend is called the load line analysis. And that's where you take the diode that you're going to be using in the circuit and you physically take it into a lab, put a voltage across it and measure the current through it. And as you increase the voltage, you create your graph to show exactly what the current is when you're at a certain voltage across that diode and then you use that in your actual circuit and you solve for things and you move things around. And it's 1) huge pain to get the data in the first place and 2) it's still a huge pain to use later. You know, depending on what you're doing, it may be necessary, probably not.
Mathematical Model
The second thing you can do is actually a mathematical model. With this math model, you can see that you have the voltage drop across the diode on the left hand of the equation, and then you have a bunch of other stuff. So you have the V S which is the source voltage across the diode, R is the resistance of the resistor that is in series with the diode. And then you have your I S , which is the current through the diode and resistor. But you also notice on the right hand of the equation, there's another V D , another voltage drops.
So this is an iterative thing, where you have to go through and not only do you have to have all of those other things figured out, you then have to iterate and try and find exactly what that voltage drop is going to be over that.
So it's a pain and again, you probably don't need it. It doesn't make a whole lot of sense, but it is one method of figuring out what that voltage drop across the diode will be if you have all of the other information anyway. So kind of dumb, but I just wanted to introduce those because there are multiple ways of approaching this. And just because there's a lot of ways doesn't mean that you should use all of them. Again, if you're going to be doing your PhD thesis, then do what you got to do.
Ideal Diode Model
But the third way that we're going to talk about this is the simplest way and in another tutorial we talked about this. And this is where we just assume that if a diode is forward biased, there is no voltage drop across it. And if it's reverse bias, then the voltage is whatever, it just happens to be in the circuit. So as we do this, there's going to be four steps that we follow one when solving a circuit with the ideal diode model.
So here, I'm going to create a circuit. I got this from taking a microelectronics class at Boise State, just pulling off there and if I were wise, I would have practiced this beforehand and we will see if it goes to complete heck in a hand basket again. And I am creating this circuit, so that we can try to go through those four steps that I just mentioned.
So the first thing just created the circuit. Now I have to look at the circuit. And wow, that is a terrible resistor that, that may be one of the worst resistors I've ever made in my entire life. So we're looking at the circuit, and we're deciding which diode is going to be forward biased and which diode is going to be reverse biased. And so I have to just kind of guess, looking at it and think, okay, I've got nine volts here. And so if these were both forward biased, I would have zero volts here, which would mean that I'd have current flowing this way that makes sense, and then I'd have a negative six volts there. But then is that being a much smaller resistor than that? Is that going to make it so I'm actually getting flow this way? Or is the flow going to be going back that way? And you just have to look at it and make that assumption and that is the very first step.
So on this one, I'm going to look at this and say, I believe that this is going to be forward so it's going to be treated as a short circuit in this ideal diode model, and then this will also be treated as a short circuit in the ideal model. So let's recreate this with those assumptions.
So we're going to go down, and that actually didn't get any better. And then we'll put a note there, show that as a short circuit to ground. And then we'll go there and show that as a short circuit to 43k and -6 volts, +9 volts, 22k, and 43k. So that's our second step is now that we've made our assumptions we've recreated the circuit.
The third step is now doing some basic circuit analysis to see if our assumptions actually work.
So in this case, we are going to have basically, two... let me do my ohm's law in the corner like I always do, just keep things straight in my head, we're going to have our two currents right here. I call this I 1 , and I 2 . And I 1 is simply nine volts minus zero volts over 22k and then I 2 is going to be zero volts minus -6 volts over 43k. And so let me throw that into a calculator really quick because there's no way I'm gonna be able to do that in my head. And that comes out to be about 409 microamps. And this comes out to be about 140 microamps. Okay, so that looks good.
Now one of the interesting things here is that I 1 is not the current that goes through this diode. It's the current that goes through the diode and also through this diode. So we're going to mark that as I 3 and we are simply going to find I 3 as I 3 equals I 1 minus I 2 . So that is going to be 409 microamps minus 140 microamps and that will give us a number that, again, I can't do in my head, 269 microamps, okay.
So now we have actually solved and we have an assumption that we are going to have zero volts across both of these diodes and we have assumed that we are going to have 140 microamps through that and 269 microamps through that way. So our quiescent point or Q-point for, let's do some retcon, for D 1 is our voltage is going to be zero volts and our current is going to be 269 microamps and D 2 … Wow, I've just I'm having one of those days, D 2 , subs of two is also going to be zero volts and 140 microamps.
Now, frankly, the the way I did this, I often see it done another way where you actually have the current first I'm used to just doing volts and amps. But you might see it where people say, hey actually do current first so 140 microamps and then zero volts in the that will be your Q-point for D 1 and your Q-point for D 2 .
Now this is where we get to our final step in the process. And this is where we look at it and say, Does this make sense? Now for there to be zero volts across these diodes, we need to have a positive current in these directions that we've established. So as we look at this and we say all right, D 1 has a positive 269 microamps in that direction and D 2 has a positive 140 microamps in that direction. And we are assuming that those are both zero volts. All of these numbers make sense with the way we've set up. So we can actually look at this and say, Yes, my assumption that both of these are forward biased is correct.
And so this is how you do it with an ideal model, when you're trying to solve for the diode, and it's pretty straightforward in terms of making it simple, just shorting that stuff. And the only difference between this in the fourth and final method that we're going to go over is that in the constant voltage drop model, we have to make things more interesting in there. You can't just put shorts or opens, well, I guess you can still put opens, but let's get into that right now.
Constant Voltage Drop Model
So let's do another circuit. So this time, we're going to start with +6 volts. So have our node right there, have our 43k right there, another node. And if we want to zoom through this, that's totally fine with me. You know, I've had people tell me that I should have been a lot of things, but an artist was never one of them. I don't even know. Was I trying to put -9? -9 volts. Okay, is that legible? I think that's legible. Okay, then we have our 22k. So now we are set up for our constant voltage drop (CVD) style.
So same exact four steps that we did in the previous one of first looking at it, making some assumptions. Second, drawing out the model with our assumptions. Third, solving for that model. And then fourth, going back, seeing if our assumptions work. And if they do great, we move forward. If not, we go back and change our assumptions and do the entire thing over again.
But the difference is with this should have pointed this out. This is very, very crucial difference between this model and the ideal model is in this one, we're going to assume that it's a 0.7 volt drop. So if we are forward biased, it's not going to just be alright, forward bias. Let's put in a short we're actually going to make model it with a voltage drop, a voltage source, of 0.7 volts. So six volts, I'm still going to make an approximation that this is going to be about zero volts. If that is forward, in reality, it'd be 0.7. And that would put 0.7, -9, I think that this is going to be forward biased and this is going to be reverse biased.
So now on step two - create that model, six volts down... why is that so bad. And then I put in my voltage source of 0.7 volts and then I put this as an open to -9 volts and 22k, step two is done.
Step three, let's actually solve for this. Now with that being gone that makes this pretty simple. So let's see if I can, in this case, I 1 will be the same thing as going through that and then we have I 2 which is going to be pretty straightforward because that's going to be zero because it's open but we model it we have I 1 . Let's see what should I do this, I 1 equals six minus 0.7 over 43k divided by 43,000. It's gonna give me 123 microamps and it is going to put this voltage right here it's going to be 0.7, and this voltage at this node is going to be -9 volts and we assume I 2 is just going to be zero because nothing can flow through that.
Okay, so now let's take this and establish our Q-points. So we are going to again retcon this because I don't do things in order like I should, D 1 one equals… Let's put this as a current first 123 microamps and 0.7 volts. And then D 2 's Q-point is going to be zero amps and -8.3… Nope, nope, I totally got that backwards because I cannot do math in my head. So that is 0.7 minus and I went the wrong direction. So this should be -9.7. Now the beautiful thing is that it doesn't change the way that it's biased. It just changes the fact that I was totally screwed up.
So now again, did everything taking our time breathing, making sure we're not making mistakes. Come on, Josh. Don't make mistakes. Now we verify that this is actually set up. So we assumed that D 1 was going to be forward biased. And we assumed D 2 was going to be reverse biased. So as we look at this, we have a positive current through it, and then that 0.7 across it just because that's what we have. And then D 2 , we have zero current through in a big -9.7 volts across it.
So both of these are correct in that we assumed this was going to be forward biased, and it is, and we assume this is going to be reversed biased, and it also is. Now again, either of these have been wrong, we'd have to go back, you'd have to say, all right, I was wrong here. Let's assume that is forward biased. And we'd have to recreate this, redo all that, which would of course, increase the chance of me being unable to do simple arithmetic again. But hopefully, that is more straightforward.
So again, the only difference between the constant voltage drop and the ideal model is the fact that you put in a voltage source to say, okay, we're losing 0.7, or whatever your assumption is, 0.7 volts across this diode. And in most cases, it won't make a difference, but on occasion it will, it definitely will make things more complicated for you.
So four different types of ways of solving four diodes in an equation, four steps in the way we do the ideal and the constant voltage drop. And despite this, despite the iteration, and despite the fact that I can't do math, this isn't that bad. Doing the constant voltage drop and doing the ideal model are pretty straightforward. So, just take your time, go through them. Make sure that at the end, you look at it and say, does it actually make sense? So hopefully that helped. If it didn't confuse you, that is fantastic. Hope you liked this, subscribe to our channel . We love people to subscribe and ask questions and learn more about electronics. We'll catch you on the next one.
Authored By
Josh bishop.
Interested in embedded systems, hiking, cooking, and reading, Josh got his bachelor's degree in Electrical Engineering from Boise State University. After a few years as a CEC Officer (Seabee) in the US Navy, Josh separated and eventually started working on CircuitBread with a bunch of awesome people. Josh currently lives in southern Idaho with his wife and four kids.
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How to Solve Parallel Circuits
Last Updated: October 27, 2022 References Approved
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 35 people, some anonymous, worked to edit and improve it over time. There are 9 references cited in this article, which can be found at the bottom of the page. wikiHow marks an article as reader-approved once it receives enough positive feedback. In this case, several readers have written to tell us that this article was helpful to them, earning it our reader-approved status. This article has been viewed 551,094 times. Learn more...
Solving parallel circuits is an easy process once you know the basic formulas and principles. When two or more resistors are connected side by side the current can "choose" it's path (in much the same way as cars tend to change lanes and drive alongside one another when a one-lane road splits into two parallel lanes). [1] X Research source After reading these steps you should be able to find the voltage, current and resistance between two or more resistors in parallel.
Cheat Sheet
- Total resistance R T for resistors in parallel: 1 / R T = 1 / R 1 + 1 / R 2 + 1 / R 3 + ...
- Voltage is always the same across branches: V T = V 1 = V 2 = V 3 = ...
- Total current I T = I 1 + I 2 + I 3 + ...
- Ohm's Law: V = IR
Introduction to Parallel Circuits
- Components "connected in parallel" are each located on a separate branch.
- For example, a circuit has two resistors in parallel, each with 4Ω resistance. 1 / R T = 1 /4Ω + 1 /4Ω → 1 / R T = 1 /2Ω → R T = 2Ω. In other words, two branches of equal resistance are exactly twice as easy to get through as one branch alone.
- If one branch has no resistance (0Ω), all the current goes through that branch. The total resistance is 0. [6] X Research source
- Make sure every value refers to the same portion of the circuit. You may use Ohm's Law to examine the total circuit (V = I T R T ) or a single branch (V = I 1 R 1 ).
Example Circuit
Additional Calculations
- R T = R 1 R 2 / (R 1 + R 2 )
- For example, two identical resistors in parallel provides ½ the total resistance of one resistor alone. Eight identical resistors provide ⅛ of the total resistance.
- Two resistors in parallel: I 1 = I T R 2 / (R 1 + R 2 )
- More than two resistors in parallel: To solve for I 1 , find the combined resistance of all resistors besides R 1 . Remember to use the formula for resistors in parallel. Now use the equation about, replacing R 2 with your answer.
Community Q&A
Video . By using this service, some information may be shared with YouTube.
- If solving Series-Parallel circuits, solve the Parallel parts first. Then you are left with a much easier Series circuit. ⧼thumbs_response⧽ Helpful 24 Not Helpful 5
- You may have been taught Ohm's Law as E = IR or V = AR. These are just different notations, meaning the same thing. ⧼thumbs_response⧽ Helpful 12 Not Helpful 5
- In a Parallel circuit the same voltage is applied across all the resistors. ⧼thumbs_response⧽ Helpful 47 Not Helpful 39
You Might Also Like
- ↑ https://www.swtc.edu/ag_power/electrical/lecture/parallel_circuits.htm
- ↑ https://www.physicsclassroom.com/class/circuits/Lesson-4/Parallel-Circuits
- ↑ https://pressbooks.bccampus.ca/basicelectricity/chapter/ground/
- ↑ http://www.tpub.com/neets/book1/chapter3/1-34.htm
- ↑ https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Book%3A_Electric_Circuits_I_-_Direct_Current_(Kuphaldt)/05%3A_Series_And_Parallel_Circuits/5.03%3A_Simple_Parallel_Circuits
- ↑ https://www.allaboutcircuits.com/textbook/direct-current/chpt-5/simple-parallel-circuits/
- ↑ https://www.allaboutcircuits.com/textbook/direct-current/chpt-5/power-calculations/
- ↑ http://physics.bu.edu/py106/notes/Circuits.html
- ↑ https://www.wilsonware.com/electronics/kirchhoff.htm
About This Article
To solve parallel circuits, you'll need to know that parallel circuits have two or more branches that all lead from point A to point B. If you want to solve for total current, use the equation IT = I1 + I2 + I3 where IT is the total current, and I1 through I3 are the currents in each branch. If you want to solve for total resistance, simply use the equation 1/RT = 1/R1 + 1/R2 + 1/R3, where "R" equals the right side. The voltage of a circuit is the same on all sides and you can use Ohm's law, or V = IR, to find missing values. To see an examples of solving these equations, scroll down. Did this summary help you? Yes No
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DC Circuits
(8 reviews)
Chad Davis, University of Oklahoma
Copyright Year: 2016
Publisher: University of Oklahoma Libraries
Language: English
Formats Available
Conditions of use.
Learn more about reviews.
Reviewed by Amr Metwally, Instructor, Oregon Institute of Technology on 4/19/20
This text book covers the basics of DC circuit analysis. It could be considered as a valuable reference book. However, I would not consider it as a text book as it lacks problems. The links provided in the book would make this book a true... read more
Comprehensiveness rating: 3 see less
This text book covers the basics of DC circuit analysis. It could be considered as a valuable reference book. However, I would not consider it as a text book as it lacks problems. The links provided in the book would make this book a true electronic book. The book has limited examples for an introductory course. Multisim examples are quite useful.
Content Accuracy rating: 5
The text book provides accurate information to the best of my knowledge.
Relevance/Longevity rating: 5
DC circuit analysis is a traditional topic in all electrical engineering curricula around the world. The future updates to this book would be adding more examples, using other simulation software and adding problems. The arrangement would easily allow those updates.
Clarity rating: 4
The book is adequately clear. A chapter summary would be so helpful to the students. In module 3, adding a section on source free circuits will be a good introduction to this chapter.
Consistency rating: 5
The textbook is consistent in terms of topics or terminology, labeling, and format. The book is also consistent with major books in this field.
Modularity rating: 5
The text is professionally divided in modules. The sub-modules well serves the understanding of their modules.
Organization/Structure/Flow rating: 5
The book is well organized. The sequence of the chapters is very logical. That well serves the understanding of the whole book.
Interface rating: 3
The links I used worked well. The hand written Appendix A1 does not look professional.
Grammatical Errors rating: 5
I could not notice any grammatical errors.
Cultural Relevance rating: 5
I could not see any cultural issues.
In general, the book is a valuable free reference book.
Reviewed by Resmi KrishnankuttyRema, Assistant Teaching Professor, Bowling Green State University on 12/12/19
This books provides both introductory and also in-depth information on the DC circuit fundamentals and associated components. It would have been really good if few more solved problems from each sections are included so the students using the book... read more
Comprehensiveness rating: 4 see less
This books provides both introductory and also in-depth information on the DC circuit fundamentals and associated components. It would have been really good if few more solved problems from each sections are included so the students using the book will get more practice problems. Including some practice questions at the end of the chapters (with or without answer key) is also something that would be useful.
I found the contents in the book accurate to the best of my knowledge.
This book talks about the fundamentals of the DC circuits in detail. I found this book relevant to the engineering curriculum.
When you look at the book first, the contents seems to be really densely packed, which might discourage a student who is really new to the topic. But when you read through the book, you can see that everything is explained well and in detail.
I found the book consistent in terms of the terminologies, topics covered, representation style and format.
Modularity rating: 4
The style of presentation of the book might be a little bit intimidating for a student in the freshman level introductory course and that might affect the modularity to some level. But I believe that if a student is willing to go through the textbook and get familiar with the layout , then it should make further navigation easy for them.
I think the book is organized well.
Interface rating: 5
The book is presented well as far as interfacing is concerned. There are plenty of circuit diagrams created using Multisim that can be easily recreated in hardware or software if needed or can be used for practicing problem solving. The hyperlinks in the text also helps with better navigation.
The text contains no grammatical errors.
The contents are purely technical. So it has no cultural relevance.
This looks like a good resource for the students to refer for DC circuits topics. The book being part of the Open Textbook Library helps the students to easily access the contents.
Reviewed by Jacob LeBlanc, Instructor, University of Louisiana at Lafayette on 11/10/19
Provides extensive information relating to DC circuits, but electromagnetism material is limited. read more
Provides extensive information relating to DC circuits, but electromagnetism material is limited.
Author clearly demonstrates mastery of subject matter in simplifying series-parallel circuits. The preface gives the impression that the author has high credibility in the subject, having had much experience of explaining “a variety of DC circuit problems in a more efficient and simple manner than you normally find in circuits books (3).”
Introductory information is clear and concise, not too intimidating for beginning electronics students. Having drawings provided from MultiSim allows readers with access to the software to be able to reconstruct the schematics and obtain the results themselves, to verify their calculations.
RC charging capacitor circuit could be more simplified.
Consistency rating: 4
It would be nice for subscripts to be used more consistently in the labeling of resistors, but it is not crucial to understanding the concepts.
The modular structure of the book is similar to how DC portions of electronics courses are normally taught—beginning with the simplest of electronic fundamentals, introducing Ohm’s Law and Power Law, then providing series circuit and parallel circuit examples to introduce Kirchhoff’s Voltage Law and Kirchhoff’s Current Law, respectively. All this being introduced in Module 1 allows for a solid background before moving into more advanced electronics problems such as those found in Modules 2 and 3.
Organization/Structure/Flow rating: 4
At the end of Modules 1 and 3 is a helpful equation list that can be referred to. Additionally, the digital bookmark links are very practical. However, the use of level headings and subsections could be better. For example, the Maximum Power Transfer Theorem should have its own section, but it is combined with Section 2.6 – Thevenin and Norton Equivalent Circuits.
At first glance, Appendix A.2 is labeled as Solving RLC circuits, but it solves an RL circuit and an RC circuit; I didn’t see any with both capacitors and inductors. This may have been intentional (referring to RL and RC circuits as RLC, but I misinterpreted). Other than that, I did not notice any issues.
Did not notice any problems.
Did not notice any problems. Author caters to beginners, intermediates, and experts alike.
Overall, Davis' DC Circuits is a great resource for those unable to pay the hundreds of dollars for conventional textbooks. An accompanying lab manual would be a great addition for future works.
Reviewed by Deborah Lichniak, Asst. Prof., EET Program Chair, Thomas Nelson Community College on 3/25/19
Although thorough the book overall seems to lend itself to those having a strong background in math. Those in a technical program are a bit intimidated by the uses of matrices and long equations and will defer to their calculators, which is not... read more
Comprehensiveness rating: 5 see less
Although thorough the book overall seems to lend itself to those having a strong background in math. Those in a technical program are a bit intimidated by the uses of matrices and long equations and will defer to their calculators, which is not presented here. I thought the material covered was the best I've seen in a while, but it did go rather quickly and there doesn't seem to be much in the way of problems students can fall back on to gauge varying degrees of comprehension. Also missing are applications where this information will be useful either now or later. Otherwise a good book to review the DC course materials in any basic electronics/electricity class. Just needs to go a bit slower. The use of MultiSim was very good as well. The most important missing bit of information I saw was that the author didn't use KVLs to complete the problems he wanted the student to find unknowns with. He relied on MultiSim and the use of a KVL or KCL appropriately would have been just as easy.
I found no errors in the book, nor any typos. It was nice to browse through a book that seems to be written by one individual instead of many.
It will last a lifetime. Already has with few exceptions.
Was a bit too wordy and seemed overwhelming to browse with the associated on-line materials. Perhaps the actual book is better. Needed to be spaced out a bit more so student can write notes with corresponding materials in case they have the book.
Stepped through all points of contention very well, though some points more quickly than others.
As I stated earlier, the ability to just browse and stay with the material is a bit daunting with all the findings cramped into a small space for a contextual look through MultiSim's eyes.
This follows the material exactly as I teach it.
I don't believe there are any problems with the graphics or links, only that they needed a bit more room to view.
I didn't find any grammatical errors.
There is no "person" involved in presenting this material to anyone. So, no cultural biases.
As I said, it's one of the best I've seen in a while. Just need more homework problems to really test the theoretical concepts and bring them to bear.
Reviewed by Mahbube Siddiki, Instructor, University of Missouri-Kansas City on 8/2/18
This text book covers succinct but resourceful topics related to DC circuit analysis for the students to learn in the first semester for a two-semester circuit analysis course covering DC and AC circuit analysis. Three modules of this textbook... read more
This text book covers succinct but resourceful topics related to DC circuit analysis for the students to learn in the first semester for a two-semester circuit analysis course covering DC and AC circuit analysis. Three modules of this textbook provide practical information spanning relevant theorems, steady and transient analysis of circuits with active and passive elements. Important external links for relevant topic is plus for the students interested to know more.
Topics covered in the book seems to be unbiased and error free.
Topics covered in the book are relevant to DC circuit analysis as the book tittle indicates. Examples used are typical and easy to understand for DC circuit analysis. As the circuit analysis theories doesn’t change much so the book will not require change of content in near future.
The book is well written, the concepts are clearly presented. Examples are worked out detail, thus easily understandable to the students
The textbook is consistent in terms of topics or terminology covered. Typical step by step method to introduce DC circuit to the students was followed.
The book is well written in a modular fashion. Topics are divided into modules and submodules.
The book is well organized, contents are presented in a logical and sequential order.
No interface issue was found.
No grammatical error was noticed.
No culturally insensitive or offensive material was found throughout the text.
Very useful textbook for an introductory circuit analysis course. Addition of more examples and problems will give students more opportunity to challenge their learning skill. Examples of some simple math operation using MATLAB can be added in appendix as well.
Reviewed by Sangho Bok, Assistant Professor, Southern Utah University on 6/19/18
The book covers the traditional topics of DC circuit analysis with the basic circuit elements such as resistors, capacitors, and inductors. Three modules are well-divided well with enough details in each module. It provides many examples with... read more
The book covers the traditional topics of DC circuit analysis with the basic circuit elements such as resistors, capacitors, and inductors. Three modules are well-divided well with enough details in each module. It provides many examples with step-by-step solutions. Hyperlinks for external resource are included so that the interested readers can have an access to more information. This book seems good for the first semester DC circuit followed by AC circuit analysis. It is not sufficient for one semester DC/AC circuit analysis setting, though.
The material for DC circuit analysis looks accurate and unbiased.
The content is relevant to DC circuit analysis. Examples are easy to follow for student to understand the analysis.
The text is clear to understand. Problem sets at the end of each module would be useful for students to practice.
The terminology of the book is consistent. The framework looks natural and is similar to most books in the field.
The modules of the book look well-organized and the sub-modules are divided properly.
The topics in DC circuit analysis are introduced in a proper order. It would be helpful to add a section in Appendix to introduce a basic concept of Multisim because the textbook uses Multisim for the circuit diagrams throughout the book.
Interface rating: 4
The book does not have a significant interface issue. Some of the texts in the figures looks smaller, though.
No grammatical error has been noticed.
There is no culturally insensitive or offensive content.
The book looks good as a textbook or a reference for DC circuit analysis. Utilizing the links throughout the book can be very useful for the readers to enhance their understanding.
Reviewed by Edwin Hou, Professor, New Jersey Institute of Technology on 5/21/18
This textbook covers the principles of DC circuit analysis with resistors, capacitors, and inductors, and the problem-solving procedures in circuit analysis. It contains traditional topics in electric circuit analysis and matrix methods for... read more
This textbook covers the principles of DC circuit analysis with resistors, capacitors, and inductors, and the problem-solving procedures in circuit analysis. It contains traditional topics in electric circuit analysis and matrix methods for solving systems of algebraic equations. The textbook is divided into three modules and it provide practical information for an introductory course in circuit analysis. The textbook contains external links for many keywords/topics so that the readers can obtain additional information if they wish to explore further.
The content of the book appears to be error-free and unbiased.
Relevance/Longevity rating: 4
The contents of the book will not be obsolete in the near future. The examples used are typical problems encountered in DC circuit analysis.
Clarity rating: 5
The textbook is clearly written. Examples are worked out in details.
The terminology used throughout the textbook is consistent.
Within each module, the topics are further divided into sections and is well organized.
The topics in circuit analysis are presented in a logical and clear fashion.
There were no interface issue with the textbook. The images/charts appear to be of correct proportion and the links I have clicked on all worked.
For the sections I examined in detail, there were no grammatical errors.
I did not see any culturally insensitive or offensive material in the textbook.
This book is useful as a textbook or as a reference in DC circuit analysis.
Reviewed by Huimin Chen, associate professor, University of New Orleans on 3/27/18
The book covers DC circuit analysis for resistive circuits and transient analysis for circuits with inductor and capacitors. It is good resource for the students to learn circuits analysis in the first semester for a two-semester circuits analysis... read more
The book covers DC circuit analysis for resistive circuits and transient analysis for circuits with inductor and capacitors. It is good resource for the students to learn circuits analysis in the first semester for a two-semester circuits analysis sequence.
The book used Multisim software to draw the circuit schematic and conduct the simulation so that students can confirm their calculations using nodal or mesh analysis and be familiar with the circuit design and synthesis. The coverage of circuit theory seems fair and balanced.
The basic circuit theory does not change much. The book is relatively easy for students to read and practice with the practice examples.
The book is clearly written. It would be nice to have a summary for each chapter indicating the important concept and circuit analysis skills that students have to master. The Thevenin's equivalent circuit model deserves more coverage in showing the circuit modeling and synthesis skills useful for studying microelectronic circuits.
The book follows a natural sequence to introduce DC circuit analysis and then the first and second order transient circuits. What might be helpful is the coverage of operational amplifier circuits (e.g., inverting and non-inverting amplifier, addition, integration, differentiation).
The book has good modular sections covering the key skill set for DC circuit analysis.
The book is well organized.
The book contains references and online resources to further the knowledge of DC circuits and analysis skills. The coverage is balanced and easy to follow.
I do not see any grammatical error.
I do not see any cultural issue.
The book is useful for engineering students to study DC circuits with Multisim software and the subsequence AC circuits seems a natural way to complete the study on linear circuit theory.
Table of Contents
PrefaceModule 1 – The Basics of DC Circuits with Resistors
- Section 1.1 – Introduction and Basic Definitions
- Section 1.1.1 - Charge vs Current
- Section 1.1.2 - Resistance Calculations – (Resistance explained in more detail in section 1.1.3)
- Section 1.1.3 - Ohm's Law: Voltage, Current, Resistance, and Conductance
- Section 1.1.4 – Power and Energy
- Section 1.2 – Combining Resistors in Parallel or Series
- Section 1.3 – Kirchhoff's Voltage Law (KVL) and Voltage Divider Rule (VDR)
- Section 1.4 – Kirchhoff's Current Law (KCL) and Current Divider Rule (CDR)
- Module 1 – Equation List
Module 2 – Advanced Topics for DC Circuits with Resistors
- Section 2.1 – Source Transformations: Thevenin and Norton Form
- Section 2.2 – Approximate Source Transformations: Adding a virtual resistor
- Section 2.2.1 - Voltage Source Approximate Transformation
- Section 2.2.2 - Current Source Approximate Transformation
- Section 2.3 – Mesh Matrix Analysis and traditional loop analysis methods
- Section 2.4 – Nodal Matrix Analysis and traditional Nodal Analysis
- Section 2.5 – Superposition: Solving a circuit by including only one source at a time
- Section 2.6 – Thevenin and Norton Equivalent Circuits
Module 3 – DC Circuits with Resistors, Capacitors, and Inductors
- Section 3.1 – Background for Capacitors
- Section 3.2 – Background for Inductors
- Section 3.3 – Combining Inductors in Parallel and/or Series
- Section 3.4 – Combining Capacitors in Parallel and/or Series
- Section 3.5 – DC Transient Analysis with RC and RL Circuits
- Section 3.5.1 – Single Loop RL and RC Charging (Store) Circuits
- Section 3.5.2 – Single Loop RL and RC Discharging (Release) Circuits
- Section 3.6 – DC Steady State Analysis with RC, RL, and RLC Circuits
- Section 3.7 – Introduction to Passive Filters
- Module 3 – Equation List
References and LinksAppendix – Dependent Sources and Laplace Transform Examples
Ancillary Material
- Submit ancillary resource
About the Book
This book covers Direct Current (DC) circuit theory and is broken up into three modules. Module 1 covers the basics for circuits that include DC sources (voltage or current) and resistors. Even though Module 1 is not very difficult, it forms the foundation for more complicated topics in modules 2 and 3 so it is important to have a firm grasp of all Module 1 topics before moving on. Module 2 covers more difficult problem solving techniques for circuits that include only DC sources and resistors. Module 3 introduces capacitors and inductors. These non-linear reactive components are analyzed in the transient and steady state regions in circuits with DC sources in Module 3. Also annexed is a two-page cheat sheet that ENGR 2431 students at University of Oklahoma can use for exams.
About the Contributors
Chad Davis received his PhD from the University of Oklahoma in 2007. Since 2008 he has been a full-time member of the ECE faculty at OU. He holds a dual discipline (electrical & mechanical) professional engineering license in the state of Oklahoma. Prior to joining the OU-ECE faculty he worked in industry at Uponor, McElroy Manufacturing, Lucent, and Celestica. His work experience ranges from electromechanical system design to automation of manufacturing and test processes.Dr. Davis is a licensed private pilot and performs research primarily in areas related to aviation. His current research at OU involves the design and development of a new GPS Ground Based Augmentation System utilizing feedback control and the design of instrumentation and data acquisition for navigational systems. Additionally, he serves as the ECE recruiting coordinator and one of the primary academic advisors for ECE students.
Contribute to this Page
The basic tools for solving DC circuit problems are Ohm's Law , the power relationship , the voltage law , and the current law . The following configurations are typical; details may be examined by clicking on the diagram for the desired circuit.
A circuit with two loops and two sources is involved enough to illustrate circuit analysis techniques. It may be analyzed by direct application of the voltage law and the current law , but some other approaches are also useful.
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Solving typical dc/dc converter application problems
CONVERSI.MAR–Conversion Devices, Inc.–SC–12– –##
Here are some simple guidelines to help ensure maximum performance and reliability
BY ANASTASIOS SIMOPOULOS and STEVE FORRESTER Conversion Devices, Inc. Brockton, MA
Dc/dc converters have evolved from simple three-terminal regulator circuits into complex electronic subassemblies that are critical in the power distribution systems of many applications. Good fundamental engineering practices must be followed in applying the converter to achieve optimum system performance and long field life. This is especially true if the application has any special requirements (like pulse loads and load sharing) or is subject to abnormal operating environments (such as elevated temperatures or moisture). Here are some dc/dc converter applications that sometimes present special problems.
Hot-pluggable applications Fault-tolerant equipment, where critical system components must always be on-line, is common in the telecommunication industry (and to a lesser degree in many medical, data processing, and data communications systems). Typically, hot-swappable (also called hot-pluggable) supplies are used for fault-tolerant systems. A hot-swappable supply (see Electronic Products, March 1993, p. 35) can be connected to, or disconnected from, a live power bus. A power converter not designed for hot plug-in may be damaged by high inrush currents if connected to a live power bus. (Inrush current is the peak current required by a dc/dc converter at turn on.) As shown in Fig. 1, the input stage of a typical dc/dc converter has an input filter that includes capacitors connected across the input terminals. At turn-on, these capacitors present a low impedance to the power line. Under normal operating conditions, the dc power bus ramps up to full power slowly, allowing the input capacitors of the dc/dc converter to fully charge without drawing excessive current. However, connecting the converter to a high input line causes it to draw high peak currents–typically lasting a few microseconds–that could reach magnitudes of 10 to 100 times the steady-state value. The amplitude and duration of these current peaks may overstress components, adversely affecting the performance and reliability of both the dc/dc converter and its power source. To avoid damage from peak currents, the user should implement a soft-start circuit. A soft-start circuit limits the rate at which power is transferred to the output stage by controlling the “on-time” of the semiconductor switch. Switching power supplies often use negative temperature coefficient (NTC) resistors to limit inrush currents caused by connection to the ac line. NTC resistors are impractical in a dc/dc converter module because of excessive heat dissipation and the lack of space. To solve the inrush problem, the user can implement a simple external inrush current limiter circuit, comprised of components as shown in Fig. 2. This circuit can control the rate at which the dc/dc input and output capacitors charge. The capacitor, C 1 , is chosen to provide the appropriate delay to eliminate both current spikes.
Hold-up time In the data-processing industry, the use of high power density dc/dc converters has increased dramatically with the acceptance of distributed power architectures. Distributed power systems carry a host of potential application-related problems, such as thermal management, input/output protection, and parallel connection problems. One overlooked problem is hold-up time, the period when a dc/dc converter remains operating within specified limits after input power is lost. More commonly specified for ac/dc converters (typically at 16 ms), the hold-up time of a dc/dc converter is normally in picoseconds. This can be a problem in data processing (or other computer-based applications) because adequate hold-up time is needed to provide an orderly system shutdown. The hold-up time of a dc/dc converter is limited by the physical size and value of the output capacitors that can be used in a small module. To increase hold-up time, the user must install an external capacitor at the converter output. The approximate value for this capacitor is given by the formula:
C out =I*( t/ V)
where t = the desired hold-up time V = the output voltage drop allowed during the hold-up time I = the output current
Unfortunately, this simple solution may not work in many applications. If the external capacitor (C out ) is too large, it may trigger the internal short-circuit protection circuits within the dc/dc converter (the capacitor presents a short to the converter output as it begins to charge). This prevents the converter from starting. Another problem is that the converter starts, but the initial ramp-up in the converter output voltage is so great that it triggers the system power-fail signal. If one or both of these conditions occurs after the installation of C out , additional circuitry is required to achieve sharp converter turn-on and extended output hold-up time. These problems can be eliminated by using the hold-up time circuit shown in Fig. 3. This circuit enables the converter to turn on normally, achieving its full output voltage levels before the external capacitor (C out ) is connected across the output terminals. The circuit will not increase turn-on time or trigger the converter's short-circuit protection feature.
Remember minimum load Most dc/dc converter application problems are neither as unique as the first two examples, nor limited to a single market segment. Often, they relate to common practices that are either poorly documented by the manufacturer or easily misunderstood by the user. One often-ignored specification is minimum load, a specification not mentioned on most dc/dc converter data sheets. Nearly all multiple output dc/dc converters require maintaining a minimum load current on the primary output so that the auxiliary outputs work properly. In a typical pulse-width-modulated converter, magnetic coupling is used to semi-regulate the auxiliary outputs from the primary output. The primary output is normally the one with the highest output current. This output controls the pulse-width-modulator (PWM) through a direct feedback loop. By controlling the PWM, the primary output, in effect, controls the auxiliary outputs. An external minimum load is required to keep the PWM feedback loop activated and deliver full rated output power to the auxiliaries. The minimum load is typically set at 10% of the full rated primary output current, but can be as high as 20% to 25% depending on the converter design. Not maintaining a minimum load causes several application problems, depending on the actual loading of each output. At best, the converter outputs lose regulation and drift out of specification. If one or more of the auxiliary outputs is heavily loaded when the primary current falls below the minimum set point, output voltage levels could rise, triggering the converter overvoltage protection circuits. If overvoltage protection is not a feature of the converter, damage to the converter or load circuitry may occur.
Opening shot: Dc/dc converters form an integral part of many power distribution systems.
Fig. 1. Most dc/dc converters have an input filter stage with capacitors that charge at a slow rate.
Fig. 2. An external inrush current limiter circuit can be added to a dc/dc converter to mitigate the effects of peak currents.
Fig. 3. A hold-up time circuit enables the converter to turn on normally, achieving its full output voltage levels before the external capacitor (C out ) is connected across the output terminals.
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DC Circuits - Problem Solving Example Problem on Ohm's Law: The Basic Circuit Question An emf source of 6.0 V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-free. What is the resistance of the lamp? Figure 1: Diagram of the circuit in this problem. Hints
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There are many ways to solve the above system. One way is to use equation (XIII) and substitute i_1 by i_2 + i_3 in equations (XI) and (XII) to obtain a system with two unknowns 100 (i_2 + i_3) + 300 i_2 = 20 (14) 300 i_2 - 50 i_3 = - 5 (15) Rearrange to rewrite the above system as 400 i_2 + 100 i_3 = 20 (16) 300 i_2 - 50 i_3 = - 5 (17)
The problem is, how do you operate an LED from a typical electronic power source, which may output 24 volts DC or more? The answer is to use a series dropping resistor: Calculate the necessary resistance value and minimum power rating of a series dropping resistor for an LED rated at 1.7 volts and 20 mA, and a power supply voltage of 24 volts.
Parallel DC Circuits Practice Worksheet With Answers Basic Electricity PDF Version Question 1 In this circuit, three resistors receive the same amount of voltage (24 volts) from a single source. Calculate the amount of current "drawn" by each resistor, as well as the amount of power dissipated by each resistor: Question 2
Step 1: Let's take stock of the circuit. It obviously only has one loop, and we've got a voltage source and two resistors. We've been given the value of the voltage source and both resistors, so all we need is to find out the current around the loop and the voltage drops over the resistors. And as soon as we find one, we can quickly use ...
DC circuit #1 See solution ↓ Circuit #2 Determine I and UAB. If U s1 and U s2 represent two ideal batteries, which one charges the other? U s1 = 120V U s2 = 90V R 1 = R 2 = 10Ω R 3 = 40Ω DC circuit #2 See solution ↓ Circuit #3 Calculate the resistance RG seen by the generator, and I1. Then, using the voltage division rule, calculate I2 and I3.
Solution : One of the typical questions in all circuit practice problems is finding the equivalent resistance of a given circuit. Recall that the equivalent of two resistors in series is R_ {eq}=R_1+R_2 Req = R1 + R2 and in parallel is \frac {1} {R_ {eq}}=\frac {1} {R_1}+\frac {1} {R_2} Req1 = R11 + R21 .
Voltages and Currents Calculator for Circuit 3. Enter the voltage source Vin in volts and the resistors R1, R2, R3, R4 and R5 in Ω and press "Calculate". The calculator uses the above formulas to calculates all currents and voltages whose formulas were obtained above. R1 =. 10.
Resistive Circuit Solver Simulate Simulate any kind of circuit you have designed and get the results instantly on just a single click. Design Design your own custom circuit using the components given in the palette in the web editor by simple drag and drop
5. Find the equivalent resistance of the circuit shown below. Find the voltage drop over, current through, and power dissipated by each resistor. Put your results in a table. We reduce circuits which are a combination of series and parallel resistors piece by piece. Examining the
Solving DC Circuit Problems Some Guidelines. Suggested Strategy Apply Kirchhoff's Rules To The Chart Method (see the 152 Web Page) Kirchhoff's Rules There are ways in which resistors can be connected so that the circuits formed cannot be reduced to a single equivalent resistor.
Circuit analysis is the process of finding all the currents and voltages in a network of connected components. We look at the basic elements used to build circuits, and find out what happens when elements are connected together into a circuit. ... DC circuit analysis. Learn. Circuit analysis overview (Opens a modal) Kirchhoff's current law ...
In this tutorial, we are going to discuss the Q-point of a diode and use few diode circuit problems to show how to solve diode circuits. We will discuss four methods of solving diode circuits: load line analysis, mathematical model, ideal diode circuit analysis, and constant voltage drop diode…
Solve for total resistance. You can find this in two different ways. You can use the resistance row to calculate it using the formula 1 / RT = 1 / R1 + 1 / R2 + 1 / R3. However, it's often easier to solve for it using Ohm's Law and the total V and I values. When solving for resistance, rearrange Ohm's Law as R = V/I Part 3 Additional Calculations 1
How do you analyze a circuit with resistors in series and parallel configurations? With the Break It Down-Build It Up Method! http://www.jesseleemason.com Mu...
This physics video tutorial explains how to solve any resistors in series and parallel combination circuit problems. The first thing you need to do is calculate the equivalent resistance of...
Module 2 covers more difficult problem solving techniques for circuits that include only DC sources and resistors. Module 3 introduces capacitors and inductors. These non-linear reactive components are analyzed in the transient and steady state regions in circuits with DC sources in Module 3.
The basic tools for solving DC circuit problems are Ohm's Law, the power relationship, the voltage law, and the current law. The following configurations are typical; details may be examined by clicking on the diagram for the desired circuit. Index. DC Circuits. HyperPhysics ***** Electricity and Magnetism.
Solve problems in DC circuits Week 3. Take the Series DC Circuits Practice Worksheet with Answers (Basic Electricity) worksheet. These questions & answers will help you master the topic! DC Circuits. A tutorial on how to use Kirchhoff's and Ohm's laws to solve DC circuit problems is presented. Examples with deatiled solutions are included.
To solve the inrush problem, the user can implement a simple external inrush current limiter circuit, comprised of components as shown in Fig. 2. This circuit can control the rate at which the dc/dc input and output capacitors charge. The capacitor, C1, is chosen to provide the appropriate delay to eliminate both current spikes. Hold-up time