AP Physics C: Electricity and Magnetism Electric Circuits — Worked Answer Explanations

Unit 3 · 12 questions explained

1. Question 1 · Easy

   A resistor $R=10\Omega$ is connected to a battery of EMF $\mathcal{E}=20\text{V}$ and internal resistance $r=2\Omega$. The current through the circuit is:

   • A
     $I=2\text{A}$

     Why not A: This uses only $R$ (ignoring internal resistance): $20/10=2\text{A}$. The total resistance is $R+r=12\Omega$.
   • B
     $I=5/3\text{A}\approx 1.67\text{A}$Correct
   • C
     $I=10\text{A}$

     Why not C: This inverts the formula, giving $R/\mathcal{E}=10/20$ instead of $\mathcal{E}/(R+r)$.
   • D
     $I=1\text{A}$

     Why not D: This uses $\mathcal{E}/(R+r+R)=20/20$, incorrectly doubling $R$ in the denominator.
   Explanation

   Applying Kirchhoff's voltage law around the single loop:
   $$\mathcal{E}-Ir-IR=0$$
   $$I=\frac{\mathcal{E}}{R+r}=\frac{20}{10+2}=\frac{20}{12}=\frac{5}{3}\text{A}\approx 1.67\text{A}$$
   The terminal voltage of the battery is $V_t=\mathcal{E}-Ir=20-(5/3)(2)=20-10/3\approx 16.7\text{V}$.

   Key takeaway

   $I = \mathcal{E}/(R+r)$; always include internal resistance in series with external load.

2. Question 2 · Easy

   Three resistors $R_1=2\Omega$, $R_2=3\Omega$, $R_3=6\Omega$ are connected in parallel across a $12\text{V}$ source. The total power delivered by the source is:

   • A
     $P=132\text{W}$

     Why not A: This may result from computing $P=V^2/R_1+V^2/R_2+V^2/R_3$ with incorrect algebra. The correct calculation gives $P=V^2/R_{\text{eq}}$.
   • B
     $P=144\text{W}$Correct
   • C
     $P=16\text{W}$

     Why not C: This uses $P=V^2/(R_1+R_2+R_3)=144/11$, treating the parallel combination as series. Parallel equivalent resistance is $1\Omega$, not $11\Omega$.
   • D
     $P=72\text{W}$

     Why not D: This uses $R_{\text{eq}}=2\Omega$ (perhaps the average of $R_1$ and $R_3$). The correct parallel equivalent is $1\Omega$.
   Explanation

   Parallel equivalent:
   $$\frac{1}{R_{\text{eq}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1{\Omega }^{-1}\Rightarrow R_{\text{eq}}=1\Omega$$

   Total power:
   $$P=\frac{V^2}{R_{\text{eq}}}=\frac{(12{)}^2}{1}=144\text{W}$$

   Alternatively: $P_1=144/2=72\text{W}$, $P_2=144/3=48\text{W}$, $P_3=144/6=24\text{W}$. Total: $72+48+24=144\text{W}$.

   Key takeaway

   Parallel: $1/R_{\text{eq}} = \sum 1/R_i$; $P = V^2/R_{\text{eq}} = IV = I^2 R_{\text{eq}}$.

3. Question 3 · Easy

   In the circuit below, ${\mathcal{E}}_1=12\text{V}$, ${\mathcal{E}}_2=6\text{V}$, $R_1=4\Omega$, and $R_2=2\Omega$. The two batteries and resistors are in a single loop, with ${\mathcal{E}}_1$ and ${\mathcal{E}}_2$ opposing each other. Applying Kirchhoff's voltage law, the current (taking ${\mathcal{E}}_1$'s direction as positive) is:

   • A
     $I=1\text{A}$Correct
   • B
     $I=3\text{A}$

     Why not B: This uses only ${\mathcal{E}}_1/(R_1+R_2)=12/6$ and ignores the opposing EMF ${\mathcal{E}}_2$. With opposing batteries, the net EMF is ${\mathcal{E}}_1-{\mathcal{E}}_2$.
   • C
     $I=-1\text{A}$

     Why not C: This assigns the wrong sign convention, treating ${\mathcal{E}}_2$ as aiding rather than opposing. With ${\mathcal{E}}_1>{\mathcal{E}}_2$, the current flows in the direction of ${\mathcal{E}}_1$.
   • D
     $I=2\text{A}$

     Why not D: This uses only ${\mathcal{E}}_1/R_1=3$ or adds EMFs: $(12+6)/6=3$ — both wrong. The net EMF is the difference for opposing batteries.
   Explanation

   KVL around the loop (clockwise, in direction of ${\mathcal{E}}_1$):
   $$+{\mathcal{E}}_1-I{R}_1-{\mathcal{E}}_2-I{R}_2=0$$
   $$({\mathcal{E}}_1-{\mathcal{E}}_2)=I(R_1+R_2)$$
   $$I=\frac{12-6}{4+2}=\frac{6}{6}=1\text{A}$$
   Positive $I$ means current flows in the direction assumed (clockwise). The ${\mathcal{E}}_2$ battery is being charged at rate $P_2={\mathcal{E}}_{2}I=6\text{W}$.

   Key takeaway

   KVL with opposing EMFs: net EMF $= \mathcal{E}_1 - \mathcal{E}_2$; current $= (\mathcal{E}_1-\mathcal{E}_2)/(R_1+R_2)$.

4. Question 4 · Easy

   An ammeter has negligible resistance. A voltmeter has very high resistance. Which connection correctly measures the voltage across resistor $R_2$ in a series circuit containing $R_1$ and $R_2$?

   • A
     Connect the voltmeter in parallel with $R_2$ and the ammeter in series in the branch containing $R_2$.Correct
   • B
     Connect both the voltmeter and ammeter in series with $R_2$.

     Why not B: A voltmeter in series has very high resistance, drastically reducing the current and giving incorrect voltage and current readings.
   • C
     Connect the voltmeter in series with $R_2$ and the ammeter in parallel with $R_2$.

     Why not C: An ammeter in parallel with $R_2$ would short-circuit $R_2$ (near-zero resistance path), and a voltmeter in series blocks current. Both connections are wrong.
   • D
     Connect both the voltmeter and ammeter in parallel with $R_2$.

     Why not D: An ammeter in parallel would short $R_2$. Ammeters must be in series to measure current without bypassing the load.
   Explanation

   Voltmeter: measures potential difference → connect in parallel across the element. High internal resistance minimizes current drawn through the meter.

   Ammeter: measures current → connect in series with the element. Near-zero resistance avoids altering the circuit current.

   This is the standard measurement configuration. In practice, placing the ammeter between the voltmeter and $R_2$ (or outside it) slightly affects readings, but for ideal meters there is no error.

   Key takeaway

   Voltmeter: in parallel (high $R$). Ammeter: in series (low $R$). Never reverse these — it damages the meters and corrupts readings.

5. Question 5 · Medium

   An RC circuit consists of a resistor $R=200\text{k}\Omega$ and capacitor $C=50\mu \text{F}$ connected in series with a battery $V_0=10\text{V}$. The switch is closed at $t=0$ with the capacitor initially uncharged. The voltage across the capacitor at time $t=\tau =RC$ is:

   • A
     $V_C(\tau )=V_0(1-e^{-1})\approx 6.32\text{V}$Correct
   • B
     $V_C(\tau )=V_0/e\approx 3.68\text{V}$

     Why not B: This is the voltage across the resistor at $t=\tau$, not the capacitor. $V_R=V_0{e}^{-t/\tau }$ while $V_C=V_0(1-e^{-t/\tau })$.
   • C
     $V_C(\tau )=V_0=10\text{V}$

     Why not C: $V_C\to V_0$ only as $t\to \infty$. At $t=\tau$, charging is only $\approx 63\%$ complete.
   • D
     $V_C(\tau )=V_0/2=5\text{V}$

     Why not D: $V_C=V_0/2$ at $t=\tau \ln 2\approx 0.693\tau$, not at $t=\tau$. The exponential function evaluated at $t/\tau =1$ gives $1-1/e\ne 1/2$.
   Explanation

   For an RC charging circuit, the ODE is:
   $$V_0=IR+V_C=RC\frac{d{V}_C}{dt}+V_C$$
   Solution with $V_C(0)=0$:
   $$V_C(t)=V_0\left(1-e^{-t/\tau }\right),\tau =RC$$
   At $t=\tau$:
   $$V_C(\tau )=V_0(1-e^{-1})=10\cdot (1-0.368)=6.32\text{V}$$
   The time constant $\tau =RC=(200\times 10^3)(50\times 10^{-6})=10\text{s}$.

   Key takeaway

   RC charging: $V_C(t) = V_0(1-e^{-t/RC})$; at $t = \tau$, capacitor is 63.2% charged.

6. Question 6 · Medium

   A fully charged capacitor $C=100\mu \text{F}$ at voltage $V_0=50\text{V}$ is discharged through resistor $R=500\Omega$ starting at $t=0$. The charge on the capacitor as a function of time is $Q(t)=Q_0{e}^{-t/\tau }$. The current through the resistor at $t=0$ is:

   • A
     $I(0)=V_0/R=0.10\text{A}$Correct
   • B
     $I(0)=Q_0/RC=0.10\text{A}$ (same value, but a different derivation — both are valid).

     Why not B: This is numerically identical to choice A (since $Q_0=C{V}_0$, so $Q_0/(RC)=C{V}_0/(RC)=V_0/R$), but in a real exam one choice would be marked correct and this would not appear as a separate choice.
   • C
     $I(0)=0$, because the current builds up from zero.

     Why not C: Zero initial current applies to an RL circuit (inductor opposes current changes). For an RC discharge, the initial current is maximum: $I(0)=V_0/R$.
   • D
     $I(0)=V_0/(R+1/C)\approx 0.10\text{A}$

     Why not D: Impedance notation ($1/C$) mixes units (Ohms vs. Farads). At $t=0$ the capacitor acts as a voltage source, so $I=V_0/R$.
   Explanation

   For the discharging capacitor, KVL gives:
   $$Q/C-IR=0\Rightarrow I=Q/(RC)=(Q_0{e}^{-t/\tau })/(RC)$$

   At $t=0$:
   $$I(0)=\frac{Q_0}{RC}=\frac{C{V}_0}{RC}=\frac{V_0}{R}=\frac{50}{500}=0.10\text{A}$$

   Current decays as $I(t)=I_0{e}^{-t/\tau }$ with $\tau =RC=(500)(100\times 10^{-6})=50\text{ms}$.

   Key takeaway

   RC discharge: $I(t) = (V_0/R)e^{-t/RC}$; initial current $= V_0/R$ (capacitor acts as voltage source at $t=0$).

7. Question 7 · Medium

   Using Kirchhoff's current law at a node, if currents $I_1=3\text{A}$ and $I_2=5\text{A}$ flow into the node, and $I_3$ flows out, which of the following is the correct application of KCL?

   • A
     $I_3=I_1+I_2=8\text{A}$Correct
   • B
     $I_3=I_1-I_2=-2\text{A}$

     Why not B: KCL requires the sum of currents into a node to equal the sum leaving. Subtraction would be appropriate only if one of these currents were directed out of the node.
   • C
     $I_3=(I_1+I_2)/2=4\text{A}$

     Why not C: KCL is not an average — it is an exact conservation law. The sum of incoming equals sum of outgoing: $I_3=I_1+I_2$.
   • D
     $I_3=0$, because current distributes evenly.

     Why not D: KCL does not imply zero net current; it requires the sum of currents at a node to be zero. Here $I_3=8\text{A}$ must leave to satisfy conservation of charge.
   Explanation

   Kirchhoff's Current Law: The algebraic sum of all currents at a node equals zero (conservation of charge).

   $$\sum I_{\text{in}}=\sum I_{\text{out}}$$
   $$I_1+I_2=I_3$$
   $$3+5=I_3=8\text{A}$$

   KCL follows from charge conservation: charge cannot accumulate at a node in a DC steady-state circuit.

   Key takeaway

   KCL: sum of currents into a node = sum out. Follows directly from conservation of charge.

8. Question 8 · Medium

   A resistor $R=50\Omega$ dissipates power $P=8\text{W}$. The current through it and the voltage across it are:

   • A
     $I=0.4\text{A}$; $V=20\text{V}$Correct
   • B
     $I=4\text{A}$; $V=200\text{V}$

     Why not B: Squaring $I$: if $I=4\text{A}$, then $P=I^{2}R=16\times 50=800\text{W}\ne 8\text{W}$. Off by a factor of 100.
   • C
     $I=0.04\text{A}$; $V=2\text{V}$

     Why not C: Checking: $P=I^{2}R=(0.04{)}^2(50)=0.08\text{W}\ne 8\text{W}$. This corresponds to $P/100$ of the given value.
   • D
     $I=0.16\text{A}$; $V=8\text{V}$

     Why not D: Checking: $P=IV=(0.16)(8)=1.28\text{W}\ne 8\text{W}$. This may come from incorrectly computing $I=P/R$ rather than $I=\sqrt{P/R}$.
   Explanation

   Using $P=I^{2}R$:
   $$I=\sqrt{\frac{P}{R}}=\sqrt{\frac{8}{50}}=\sqrt{0.16}=0.4\text{A}$$
   Voltage: $V=IR=0.4\times 50=20\text{V}$.

   Verify: $P=IV=(0.4)(20)=8\text{W}$ ✓

   Alternately from $P=V^2/R$: $V=\sqrt{PR}=\sqrt{8\times 50}=\sqrt{400}=20\text{V}$.

   Key takeaway

   Power: $P = I^2R = V^2/R = IV$. Use $I = \sqrt{P/R}$ or $V = \sqrt{PR}$ when power and resistance are given.

9. Question 9 · Medium

   In a circuit with two parallel resistors $R_1=6\Omega$ and $R_2=3\Omega$ in series with $R_3=4\Omega$, connected to a $12\text{V}$ ideal battery, the voltage across the parallel combination is:

   • A
     $V_{\parallel }=4\text{V}$Correct
   • B
     $V_{\parallel }=6\text{V}$

     Why not B: Checking: total resistance would be $3+4=7\Omega$, current $=12/7\approx 1.71\text{A}$, voltage across parallel $=1.71\times 2=3.43\text{V}\ne 6\text{V}$. The parallel combination is $2\Omega$, not $3\Omega$.
   • C
     $V_{\parallel }=8\text{V}$

     Why not C: This is the voltage across $R_3=4\Omega$ (not the parallel part). The total current is $2\text{A}$, giving $V_3=2\times 4=8\text{V}$ and $V_p=12-8=4\text{V}$.
   • D
     $V_{\parallel }=12\text{V}$

     Why not D: The full battery voltage appears only across the complete series-parallel network, not across just the parallel portion. Series element $R_3$ drops $8\text{V}$, leaving $4\text{V}$ for the parallel combination.
   Explanation

   Parallel combination: $1/R_p=1/6+1/3=(1+2)/6=1/2$, so $R_p=2\Omega$.

   Total resistance: $R_p+R_3=2+4=6\Omega$.

   Total current: $I=12/6=2\text{A}$.

   Voltage across parallel part: $V_p=I\cdot R_p=2\times 2=4\text{V}$.

   Check: $V_{R_3}=2\times 4=8\text{V}$; $4+8=12\text{V}$ ✓

   Key takeaway

   Series-parallel: find equivalent $R_p$, compute total $I$, then $V_p = I\cdot R_p$.

10. Question 10 · Hard

    An RC circuit has $R=10\text{k}\Omega$, $C=20\mu \text{F}$, and battery $V_0=6\text{V}$. The switch is closed at $t=0$ with $C$ initially uncharged. The energy dissipated in the resistor from $t=0$ to $t\to \infty$ is:

    • A
      $E_R=\frac{1}{2}C{V}_0^2=360\mu \text{J}$Correct
    • B
      $E_R=C{V}_0^2=720\mu \text{J}$

      Why not B: This equals the total energy delivered by the battery ($C{V}_0^2$), not the energy stored in the capacitor or dissipated in $R$. The battery delivers $C{V}_0^2$ total: half stored in $C$, half dissipated in $R$.
    • C
      $E_R=\frac{V_0^2}{2R}=1.8\times 10^{-3}\text{J}$

      Why not C: This has units of $\text{W}$ ($=V^2/(2R)$), not joules (energy). Power integrated over infinite time diverges; the finite result comes from the capacitor limiting total charge flow.
    • D
      $E_R=0$, because an ideal resistor stores no energy.

      Why not D: While a resistor stores no potential energy, it does dissipate energy as heat. The total dissipation equals $\frac{1}{2}C{V}_0^2$.
    Explanation

    Total energy delivered by battery (charge $Q_{\infty }=C{V}_0$ at voltage $V_0$):
    $$E_{\text{battery}}=Q_{\infty }{V}_0=C{V}_0^2=(20\times 10^{-6})(36)=720\mu \text{J}$$
    Energy stored in capacitor:
    $$U_C=\frac{1}{2}C{V}_0^2=360\mu \text{J}$$
    Energy dissipated in resistor (by conservation):
    $$E_R=E_{\text{battery}}-U_C=720-360=360\mu \text{J}=\frac{1}{2}C{V}_0^2$$

    This result is independent of $R$ — the same fraction of energy is always dissipated regardless of resistance. Direct integral verification:
    $$E_R={\int }_0^{\infty }{I}^{2}Rdt={\int }_0^{\infty }\frac{V_0^2}{R}{e}^{-2t/RC}dt=\frac{V_0^2}{R}\cdot \frac{RC}{2}=\frac{1}{2}C{V}_0^2$$

    Key takeaway

    RC charging: half the battery energy stored in $C$, half dissipated in $R$, always. Result is independent of $R$.

11. Question 11 · Hard

    In the Wheatstone bridge circuit, resistors $R_1$ (top-left) and $R_2$ (top-right) form one pair of arms; $R_3$ (bottom-left) and $R_X$ (bottom-right) form the other. A galvanometer connects the midpoints. The bridge is balanced when:

    • A
      $R_1/R_3=R_2/R_X$Correct
    • B
      $R_1+R_X=R_2+R_3$

      Why not B: The balance condition is a ratio (cross-product), not a sum. Equal sums do not guarantee equal midpoint potentials.
    • C
      $R_1{R}_2=R_3{R}_X$

      Why not C: This multiplies adjacent arms rather than opposite arms. The correct condition pairs opposite resistors: $R_1{R}_X=R_2{R}_3$, equivalent to $R_1/R_3=R_2/R_X$.
    • D
      $R_1/R_2=R_X/R_3$

      Why not D: This cross-multiplies to $R_1{R}_3=R_2{R}_X$, which is a different condition from the correct $R_1{R}_X=R_2{R}_3$.
    Explanation

    At balance, no current flows through the galvanometer, so both midpoints are at the same potential.

    Left divider voltage at midpoint: $V_m=V_s\cdot R_3/(R_1+R_3)$
    Right divider voltage at midpoint: $V_m=V_s\cdot R_X/(R_2+R_X)$

    Setting equal:
    $$\frac{R_3}{R_1+R_3}=\frac{R_X}{R_2+R_X}\Rightarrow R_3(R_2+R_X)=R_X(R_1+R_3)$$
    $$R_2{R}_3=R_1{R}_X\Rightarrow \frac{R_1}{R_3}=\frac{R_2}{R_X}$$
    Unknown: $R_X=R_2{R}_3/R_1$.

    Key takeaway

    Wheatstone bridge balance: $R_1/R_3 = R_2/R_X$ (ratios of opposite arms equal). Measures unknown $R_X$ precisely.

12. Question 12 · Hard

    An RC circuit with $R=1\text{k}\Omega$ and $C=10\mu \text{F}$ is driven by a square-wave voltage that steps from $0$ to $V_0=10\text{V}$ at $t=0$ (with $C$ initially uncharged). The voltage across the capacitor at $t=2\tau$ (where $\tau =RC$) is:

    • A
      $V_C(2\tau )=V_0(1-e^{-2})\approx 8.65\text{V}$Correct
    • B
      $V_C(2\tau )=V_0(1-e^{-1/2})\approx 3.94\text{V}$

      Why not B: This evaluates $V_C$ at $t=\tau /2$, not $t=2\tau$. In the exponent, $t/\tau =2$ at $t=2\tau$, not $1/2$.
    • C
      $V_C(2\tau )=2V_0(1-e^{-1})\approx 12.6\text{V}$

      Why not C: This incorrectly multiplies by 2 or confuses the driving voltage with $2V_0$. The asymptote is $V_0$, never $2V_0$ — the capacitor cannot charge above the supply.
    • D
      $V_C(2\tau )=V_0{e}^{-2}\approx 1.35\text{V}$

      Why not D: This is the charging solution only for the resistor voltage $V_R=V_0{e}^{-t/\tau }$, not the capacitor voltage. $V_C$ approaches $V_0$, while $V_R$ decays to zero.
    Explanation

    The charging ODE $V_0=RCd{V}_C/dt+V_C$ has solution:
    $$V_C(t)=V_0(1-e^{-t/\tau })$$
    with $\tau =RC=(10^3)(10\times 10^{-6})=10\text{ms}$.

    At $t=2\tau$:
    $$V_C(2\tau )=V_0(1-e^{-2})=10(1-0.1353)=10(0.8647)\approx 8.65\text{V}$$

    At $t=\tau$: $V_C\approx 6.32\text{V}$. At $t=2\tau$: $V_C\approx 8.65\text{V}$. At $t=5\tau$: $V_C\approx 9.93\text{V}$ (essentially fully charged).

    Key takeaway

    RC charging: $V_C(t) = V_0(1-e^{-t/\tau})$; at $t = 2\tau$, the capacitor is 86.5% charged.