AP Physics 2 Electric Circuits — Worked Answer Explanations

Unit 3 · 18% of the AP exam · 12 questions explained

1. Question 1 · Easy

   A resistor has a current of $I=3\mathrm{A}$ flowing through it and a voltage of $V=12\mathrm{V}$ across it. What is its resistance?

   • A
     $R=0.25\Omega$

     Why not A: This inverts the formula: $R=I/V$ instead of $R=V/I$.
   • B
     $R=4\Omega$Correct
   • C
     $R=36\Omega$

     Why not C: This multiplies $V\times I$ instead of dividing $V/I$.
   • D
     $R=9\Omega$

     Why not D: This computes $V/I^2$ or another incorrect combination.
   Explanation

   Ohm's law: $V=IR$, so $R=V/I$.
   $$R=\frac{12\mathrm{V}}{3\mathrm{A}}=4\Omega$$
   Ohm's law relates the three fundamental circuit quantities. The unit of resistance, the ohm ($\Omega$), equals $V/A$.

   Key takeaway

   Ohm's law: $V = IR$; resistance $R = V/I$ in ohms when $V$ is in volts and $I$ in amperes.

2. Question 2 · Easy

   Which of the following best describes what current represents in an electric circuit?

   • A
     The energy per unit charge delivered by the source.

     Why not A: Energy per unit charge is voltage (electric potential difference), not current.
   • B
     The rate at which charge flows past a cross section of a conductor.Correct
   • C
     The resistance of the conductor to charge flow.

     Why not C: Resistance opposes current flow; it is not the same as current.
   • D
     The power dissipated per unit resistance.

     Why not D: This does not correspond to a standard definition; power is $P=I^{2}R$ or $P=IV$.
   Explanation

   Electric current $I$ is defined as the rate of charge flow:
   $$I=\frac{\Delta Q}{\Delta t}$$
   The SI unit is the ampere (A), where $1\mathrm{A}=1C/s$. By convention, current direction is the direction of positive charge flow — opposite to electron flow in a metal. Voltage is energy per charge ($V=U/q$), and resistance is the ratio of voltage to current ($R=V/I$).

   Key takeaway

   Current $I = \Delta Q / \Delta t$; it measures charge flow rate in coulombs per second (amperes).

3. Question 3 · Easy

   Two resistors, $R_1=6\Omega$ and $R_2=3\Omega$, are connected in parallel across a battery. What is the equivalent resistance of the combination?

   • A
     $R_{\text{eq}}=9\Omega$

     Why not A: Adding resistances directly gives the series combination, not parallel.
   • B
     $R_{\text{eq}}=4.5\Omega$

     Why not B: This is the arithmetic average, not the correct parallel formula.
   • C
     $R_{\text{eq}}=2\Omega$Correct
   • D
     $R_{\text{eq}}=1\Omega$

     Why not D: This results from adding $1/R_1+1/R_2=1/6+1/3=1/2$ and taking the sum rather than its reciprocal.
   Explanation

   For parallel resistors:
   $$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}$$
   $$R_{\text{eq}}=2\Omega$$

   For two resistors in parallel, the product-over-sum shortcut also works:
   $$R_{\text{eq}}=\frac{R_1{R}_2}{R_1+R_2}=\frac{6\times 3}{6+3}=\frac{18}{9}=2\Omega$$

   The parallel equivalent is always less than the smallest individual resistor (here, less than $3\Omega$).

   Key takeaway

   Parallel: $1/R_{\text{eq}} = 1/R_1 + 1/R_2$; result is always less than the smallest resistor.

4. Question 4 · Easy

   A $12\Omega$ resistor and an $8\Omega$ resistor are connected in series to a $20\mathrm{V}$ battery with negligible internal resistance. What is the voltage across the $8\Omega$ resistor?

   • A
     $V_8=12\mathrm{V}$

     Why not A: This is the voltage across the $12\Omega$ resistor, not the $8\Omega$ one.
   • B
     $V_8=8\mathrm{V}$Correct
   • C
     $V_8=10\mathrm{V}$

     Why not C: This would apply if the voltage split equally, but voltage divides in proportion to resistance.
   • D
     $V_8=20\mathrm{V}$

     Why not D: The full battery voltage is across the entire series combination, not just one resistor.
   Explanation

   In a series circuit, the same current flows through all components. Total resistance:
   $$R_{\text{total}}=12+8=20\Omega$$
   Current: $I=V/R=20\mathrm{V}/20\Omega =1\mathrm{A}$

   Voltage across $8\Omega$:
   $$V_8=IR=1\mathrm{A}\times 8\Omega =8\mathrm{V}$$

   Check: $V_{12}=1\times 12=12\mathrm{V}$, and $8+12=20\mathrm{V}$ ✓ (KVL). Voltages in series divide in proportion to resistance.

   Key takeaway

   Series circuit: same current everywhere; voltage divides in proportion to resistance ($V_n = IR_n$).

5. Question 5 · Medium

   A circuit has a $9\mathrm{V}$ battery with internal resistance $r=1\Omega$ connected to an external resistance $R=8\Omega$. What is the terminal voltage of the battery?

   • A
     $V_{\text{terminal}}=9\mathrm{V}$

     Why not A: The terminal voltage equals the EMF only when no current flows (open circuit). Under load, internal resistance causes a voltage drop.
   • B
     $V_{\text{terminal}}=8\mathrm{V}$Correct
   • C
     $V_{\text{terminal}}=1\mathrm{V}$

     Why not C: This is the voltage dropped across the internal resistance ($Ir=1\times 1$), not the terminal voltage.
   • D
     $V_{\text{terminal}}=9.1\mathrm{V}$

     Why not D: Terminal voltage is always less than or equal to EMF when current flows; internal resistance causes a drop, not an increase.
   Explanation

   The terminal voltage accounts for the voltage drop across the battery's internal resistance:
   $$V_{\text{terminal}}=\mathcal{E}-Ir$$

   First find the current:
   $$I=\frac{\mathcal{E}}{R+r}=\frac{9}{8+1}=1\mathrm{A}$$

   Then terminal voltage:
   $$V_{\text{terminal}}=9-(1)(1)=8\mathrm{V}$$

   Note: $V_{\text{terminal}}=IR=1\times 8=8\mathrm{V}$ ✓ — the terminal voltage equals the voltage across the external resistance.

   Key takeaway

   Terminal voltage: $V_T = \mathcal{E} - Ir$; internal resistance causes a voltage drop under load.

6. Question 6 · Medium

   A $6\Omega$ resistor and a $3\Omega$ resistor are connected in parallel. This parallel combination is in series with a $2\Omega$ resistor, and the whole circuit is powered by a $12\mathrm{V}$ battery. What is the current through the $6\Omega$ resistor?

   • A
     $I_6=3\mathrm{A}$

     Why not A: This is the total current from the battery, not the branch current through the $6\Omega$ resistor alone.
   • B
     $I_6=1\mathrm{A}$Correct
   • C
     $I_6=2\mathrm{A}$

     Why not C: This is the current through the $3\Omega$ branch (smaller resistance gets more current). The $6\Omega$ branch carries half as much current as the $3\Omega$ branch.
   • D
     $I_6=4\mathrm{A}$

     Why not D: This is $12\mathrm{V}/3\Omega$ — applying the battery voltage across the wrong element; the $3\Omega$ is a branch, and the battery voltage is split between the series $2\Omega$ and the parallel combination.
   Explanation

   Step 1: Combine the parallel resistors:
   $$R_{\text{par}}=\frac{6\times 3}{6+3}=\frac{18}{9}=2\Omega$$

   Step 2: Total circuit resistance:
   $$R_{\text{total}}=R_{\text{series}}+R_{\text{par}}=2+2=4\Omega$$

   Step 3: Total current from battery:
   $$I_{\text{total}}=\frac{V}{R_{\text{total}}}=\frac{12}{4}=3\mathrm{A}$$

   Step 4: Voltage across the parallel combination:
   $$V_{\text{par}}=I_{\text{total}}\times R_{\text{par}}=3\times 2=6\mathrm{V}$$

   Step 5: Current through the $6\Omega$ branch:
   $$I_6=\frac{V_{\text{par}}}{6}=\frac{6}{6}=1\mathrm{A}$$

   Hmm — rechecking with current divider: $I_6=I_{\text{total}}\times \frac{R_3}{R_3+R_6}=3\times \frac{3}{9}=1\mathrm{A}$. And $I_3=3\times \frac{6}{9}=2\mathrm{A}$. Check: $1+2=3\mathrm{A}$ ✓.

   Correct answer is $I_6=1\mathrm{A}$ (current divider: larger resistance gets smaller fraction of current). The current divider rule: parallel branches share voltage equally; each branch current = $V_{\text{par}}/R_{\text{branch}}$.

   Key takeaway

   Mixed circuit: reduce series/parallel combinations step by step, find voltage across parallel section, then divide by branch resistance.

7. Question 7 · Medium

   A $60\mathrm{W}$ lightbulb is designed to operate at $120\mathrm{V}$. What is its operating resistance?

   • A
     $R=0.5\Omega$

     Why not A: This inverts the relationship: $P/V^2=60/14400$ is $1/R$, but here the result is in ${\Omega }^{-1}$, not $\Omega$.
   • B
     $R=240\Omega$Correct
   • C
     $R=2\Omega$

     Why not C: This computes $P/V=60/120=0.5$, which is the current $I$, not the resistance.
   • D
     $R=7200\Omega$

     Why not D: This computes $V\times P=120\times 60$, which has no physical meaning.
   Explanation

   Using the power formula $P=V^2/R$:
   $$R=\frac{V^2}{P}=\frac{(120{)}^2}{60}=\frac{14400}{60}=240\Omega$$

   Alternative: first find current $I=P/V=60/120=0.5\mathrm{A}$, then $R=V/I=120/0.5=240\Omega$ ✓

   Note: the resistance of a real bulb varies with temperature (tungsten's resistivity increases with heat), so the cold resistance is much lower than the operating resistance.

   Key takeaway

   Power formulas: $P = IV = I^2R = V^2/R$; use $R = V^2/P$ when voltage and power are known.

8. Question 8 · Medium

   Kirchhoff's voltage law (KVL) states that the sum of all voltage drops around any closed loop in a circuit equals zero. This law is a consequence of which fundamental principle?

   • A
     Conservation of charge

     Why not A: Conservation of charge underlies Kirchhoff's current law (KCL), not KVL.
   • B
     Conservation of energyCorrect
   • C
     Newton's third law

     Why not C: Newton's third law concerns action-reaction force pairs, not circuit behavior.
   • D
     Ohm's law

     Why not D: Ohm's law defines resistance for a resistor; KVL is a broader energy conservation principle that does not require Ohm's law.
   Explanation

   KVL ($\sum V=0$ around a closed loop) reflects conservation of energy: the electric potential is a well-defined function of position, so going around any closed loop must return to the same potential — net change is zero. Equivalently, a charge cannot gain or lose net energy traversing a closed path in a conservative electric field.

   KCL ($\sum I_{\text{in}}=\sum I_{\text{out}}$ at a junction) reflects conservation of charge: charge cannot accumulate at a steady-state circuit node.

   Key takeaway

   KVL → conservation of energy (potential is path-independent); KCL → conservation of charge (no charge pile-up).

9. Question 9 · Hard

   An RC circuit has a capacitor $C=200\mu \mathrm{F}$ in series with a resistor $R=5\mathrm{k}\Omega$ and a $12\mathrm{V}$ battery. The capacitor is initially uncharged. Approximately how long after the switch is closed does the voltage across the capacitor reach $7.6\mathrm{V}$? (Use $e\approx 2.718$, so $1-e^{-1}\approx 0.632$.)

   • A
     $t\approx 0.5\mathrm{s}$

     Why not A: This is half the time constant but $7.6\mathrm{V}$ corresponds to one time constant, not half.
   • B
     $t\approx 1.0\mathrm{s}$Correct
   • C
     $t\approx 2.0\mathrm{s}$

     Why not C: At $t=2\tau$, $V_C=12(1-e^{-2})\approx 10.4\mathrm{V}$, not $7.6\mathrm{V}$.
   • D
     $t\approx 0.1\mathrm{s}$

     Why not D: This would correspond to about $\tau /10$; at such a short time the capacitor voltage is much less than $7.6\mathrm{V}$.
   Explanation

   The time constant for this RC circuit:
   $$\tau =RC=(5\times 10^3\Omega )(200\times 10^{-6}\mathrm{F})=1.0\mathrm{s}$$

   Voltage across capacitor during charging:
   $$V_C(t)=V_0\left(1-e^{-t/\tau }\right)=12\left(1-e^{-t/1.0}\right)$$

   Setting $V_C=7.6\mathrm{V}$:
   $$7.6=12(1-e^{-t})\Rightarrow e^{-t}=1-\frac{7.6}{12}=1-0.633\approx 0.368=e^{-1}$$

   So $t\approx 1\tau =1.0\mathrm{s}$. After one time constant, the capacitor charges to $63.2\%$ of the supply voltage: $0.632\times 12=7.58\approx 7.6\mathrm{V}$ ✓

   Key takeaway

   RC time constant $\tau = RC$; after time $\tau$, capacitor charges to $\approx 63.2\%$ of final voltage.

10. Question 10 · Hard

    In a circuit, two batteries are connected in the same loop: Battery 1 has EMF ${\mathcal{E}}_1=12\mathrm{V}$ with internal resistance $r_1=1\Omega$, and Battery 2 has EMF ${\mathcal{E}}_2=6\mathrm{V}$ with internal resistance $r_2=2\Omega$. They are connected with opposing polarity (positive terminals facing each other). An external resistance $R=3\Omega$ completes the loop. What is the current in the loop?

    • A
      $I=3\mathrm{A}$

      Why not A: This adds the EMFs ($12+6=18\mathrm{V}$) instead of subtracting for opposing polarity.
    • B
      $I=1\mathrm{A}$Correct
    • C
      $I=2\mathrm{A}$

      Why not C: This uses only the external resistance in the denominator, ignoring the internal resistances of both batteries.
    • D
      $I=0\mathrm{A}$

      Why not D: Zero current would require the net EMF to be zero; here $12\ne 6$, so there is a non-zero net EMF driving current.
    Explanation

    Applying KVL with opposing-polarity batteries (net EMF = difference):
    $${\mathcal{E}}_{\text{net}}={\mathcal{E}}_1-{\mathcal{E}}_2=12-6=6\mathrm{V}$$

    Total resistance:
    $$R_{\text{total}}=r_1+r_2+R=1+2+3=6\Omega$$

    Current (in the direction favored by the larger EMF):
    $$I=\frac{{\mathcal{E}}_{\text{net}}}{R_{\text{total}}}=\frac{6}{6}=1\mathrm{A}$$

    KVL check around the loop: $+12-I(1)-6-I(2)-I(3)=12-1-6-2-3=0$ ✓

    Key takeaway

    Opposing batteries: net EMF = difference; apply KVL with consistent sign convention to find current.

11. Question 11 · Hard

    A fully charged capacitor ($C=100\mu \mathrm{F}$, $V_0=10\mathrm{V}$) is connected in series with a resistor $R=2\mathrm{k}\Omega$ with no battery. After the switch is closed, what is the current through the resistor at $t=0$ (immediately after closing)?

    • A
      $I_0=0$

      Why not A: At $t=0$, the capacitor acts as a voltage source; current is not zero initially.
    • B
      $I_0=5\mathrm{mA}$Correct
    • C
      $I_0=50\mathrm{mA}$

      Why not C: This uses $R=200\Omega$ instead of $2000\Omega$, a factor-of-10 error.
    • D
      $I_0=1\mathrm{A}$

      Why not D: This ignores the resistance entirely, computing $Q_0/C=V_0$, not $V_0/R$.
    Explanation

    During capacitor discharge, at $t=0$ the capacitor has full voltage $V_0=10\mathrm{V}$. It drives current through the resistor just like a battery would:
    $$I_0=\frac{V_0}{R}=\frac{10\mathrm{V}}{2000\Omega }=5\times 10^{-3}\mathrm{A}=5\mathrm{mA}$$

    As time progresses, the voltage and current decay exponentially:
    $$I(t)=I_0{e}^{-t/\tau },\tau =RC=(2000)(100\times 10^{-6})=0.2\mathrm{s}$$

    At $t=0$, the capacitor acts as a fully charged battery; as it discharges, current and voltage decay toward zero.

    Key takeaway

    At $t=0$, a capacitor acts as a voltage source; initial discharge current is $I_0 = V_0/R$.

12. Question 12 · Hard

    Three resistors are connected between nodes A, B, and C: $R_{AB}=3\Omega$ (directly between A and B), $R_{BC}=3\Omega$ (between B and C), and $R_{AC}=6\Omega$ (between A and C). A $12\mathrm{V}$ battery is connected between A and B. What is the total current delivered by the battery?

    • A
      $I=4\mathrm{A}$

      Why not A: This uses only $R_{AB}=3\Omega$ as if the other resistors are open, ignoring the parallel path A→C→B.
    • B
      $I=5.33\mathrm{A}$Correct
    • C
      $I=1\mathrm{A}$

      Why not C: This treats all three resistors as series ($3+3+6=12\Omega$), but $R_{AB}$ is in parallel with the series path $R_{AC}+R_{BC}$.
    • D
      $I=2.67\mathrm{A}$

      Why not D: This uses $R_{AC}$ and $R_{BC}$ in parallel ($=2\Omega$) instead of series ($=9\Omega$) before combining with $R_{AB}$.
    Explanation

    The battery drives current from A to B. Two parallel paths exist:

    • Path 1 (direct): A → B through $R_{AB}=3\Omega$
    • Path 2 (indirect): A → C (through $R_{AC}=6\Omega$) → B (through $R_{BC}=3\Omega$): total $=9\Omega$

    Equivalent resistance of the two parallel paths:
    $$\frac{1}{R_{\text{eq}}}=\frac{1}{3}+\frac{1}{9}=\frac{3}{9}+\frac{1}{9}=\frac{4}{9}\Rightarrow R_{\text{eq}}=\frac{9}{4}=2.25\Omega$$

    Total current from battery:
    $$I=\frac{V}{R_{\text{eq}}}=\frac{12}{9/4}=12\times \frac{4}{9}=\frac{48}{9}\approx 5.33\mathrm{A}$$

    Branch currents: $I_1=12/3=4\mathrm{A}$ (direct path), $I_2=12/9=1.33\mathrm{A}$ (indirect path). Check: $4+1.33=5.33\mathrm{A}$ ✓

    Key takeaway

    Network between two nodes: find all paths, treat as parallel; $R_{\text{eq}}$ from all parallel branch resistances combined.