Practice 2: Electric Circuits

10 MCQ + 5 FRQ · Covers Ohm's Law, Kirchhoff's Rules, RC Circuits, Power, and Resistivity

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Multiple Choice Questions

Q1. A light bulb rated at 100 W operates at 120 V. What is its resistance when lit?

Correct! \(P = V^2/R\), so \(R = V^2/P = (120)^2/100 = 14400/100 = 144\ \Omega\).
Incorrect. The correct answer is (C) \(144\ \Omega\). Using \(P = V^2/R\): \(R = (120)^2/100 = 144\ \Omega\).

Q2. A computer draws 0.5 A at 120 V for 8 hours per day. If electricity costs $0.12 per kWh, what is the monthly (30 day) cost?

Correct! Power: \(P = IV = 0.5 \times 120 = 60\ \text{W} = 0.06\ \text{kW}\). Daily energy: \(0.06 \times 8 = 0.48\ \text{kWh}\). Monthly: \(0.48 \times 30 = 14.4\ \text{kWh}\). Cost: \(14.4 \times \$0.12 = \$1.73\).
Incorrect. The correct answer is (D) $1.73. \(P = 0.5 \times 120 = 60\ \text{W}\). Monthly kWh: \(0.06 \times 8 \times 30 = 14.4\ \text{kWh}\). Cost: \(14.4 \times \$0.12 = \$1.73\).

Q3. How many electrons pass through a resistor in 2.0 s if the current is 3.0 A? (\(e = 1.6 \times 10^{-19}\ \text{C}\))

Correct! \(Q = It = 3.0 \times 2.0 = 6.0\ \text{C}\). \(N = Q/e = 6.0/(1.6 \times 10^{-19}) = 3.75 \times 10^{19}\).
Incorrect. The correct answer is (A) \(3.75 \times 10^{19}\). \(Q = It = 6.0\ \text{C}\), \(N = Q/e = 3.75 \times 10^{19}\).

Q4. Two identical resistors are connected first in series, then in parallel, to the same battery. The ratio of total power in parallel to total power in series is:

Correct! Series: \(R_{\text{eq}} = 2R\), \(P_s = V^2/(2R)\). Parallel: \(R_{\text{eq}} = R/2\), \(P_p = V^2/(R/2) = 2V^2/R\). Ratio: \(P_p/P_s = (2V^2/R)/(V^2/2R) = 4\).
Incorrect. The correct answer is (B) \(4\). Series: \(R_{\text{eq}} = 2R\). Parallel: \(R_{\text{eq}} = R/2\). Ratio = \((V^2/(R/2))/(V^2/(2R)) = 4\).

Q5. Bulb A is in series with the parallel combination of bulbs B and C. Switch S shorts out bulb C when closed. When S is closed, what happens to bulb A?

Correct! With S open: \(R_{\text{eq}} = R + (R \parallel R) = 1.5R\). Current = \(V/1.5R\). With S closed: C shorted, \(R_{\text{eq}} = R + 0 = R\). Current increases to \(V/R\). Bulb A gets brighter.
Incorrect. The correct answer is (C). Closing S shorts bulb C, reducing \(R_{\text{eq}}\) from \(1.5R\) to \(R\). More current flows, so bulb A gets brighter.

Q6. A tungsten wire has resistance \(R_0\) at 20°C with \(\alpha = 4.5 \times 10^{-3}\ \text{°C}^{-1}\). At what temperature is its resistance \(2R_0\)?

Correct! \(R = R_0[1 + \alpha(T - T_0)]\). \(2 = 1 + 4.5\times10^{-3}(T-20)\). \(T-20 = 1/(4.5\times10^{-3}) = 222\). \(T = 242\ \text{°C}\).
Incorrect. The correct answer is (D) 242 °C. \(2 = 1 + \alpha(T-20)\), so \(T-20 = 1/\alpha = 222\), \(T = 242\ \text{°C}\).

Q7. Two wires have equal length and resistance. Wire X is aluminum (\(\rho_{\text{Al}} = 2.82\times10^{-8}\ \Omega\cdot\text{m}\)), wire Y is copper (\(\rho_{\text{Cu}} = 1.72\times10^{-8}\ \Omega\cdot\text{m}\)). What is \(d_X/d_Y\)?

Correct! \(R = \rho L/A = 4\rho L/(\pi d^2)\). For equal \(R\) and \(L\): \(d \propto \sqrt{\rho}\). \(d_X/d_Y = \sqrt{2.82\times10^{-8}/1.72\times10^{-8}} = \sqrt{1.64} = 1.28\).
Incorrect. The correct answer is (D) 1.28. \(d \propto \sqrt{\rho}\) for equal \(R, L\). \(d_X/d_Y = \sqrt{2.82/1.72} = 1.28\).

Q8. Unknown resistor \(X\) in parallel with \(Y = 20\ \Omega\) gives \(R_{\text{eq}} = 15\ \Omega\). Find \(X\).

Correct! \(1/15 = 1/X + 1/20\) → \(1/X = 1/15 - 1/20 = (4-3)/60 = 1/60\) → \(X = 60\ \Omega\).
Incorrect. The correct answer is (B) \(60\ \Omega\). \(1/15 = 1/X + 1/20\) → \(1/X = 1/15 - 1/20 = 1/60\) → \(X = 60\ \Omega\).

Q9. In an RC circuit, the capacitor is initially uncharged. At \(t = 0\), the switch is closed. How does the current through the resistor change with time?

Correct! Charging RC: \(I(t) = (\mathcal{E}/R)e^{-t/RC}\). Current starts at \(\mathcal{E}/R\) and decays exponentially to zero as the capacitor charges.
Incorrect. The correct answer is (E). \(I(t) = (\mathcal{E}/R)e^{-t/RC}\), decaying exponentially from max to zero.

Q10. A capacitor \(C\) charged to \(V_0\) discharges through \(R\). After \(t = RC\), the energy stored is what fraction of its initial energy?

Correct! Discharging: \(V(t) = V_0 e^{-t/RC}\). At \(t = RC\): \(V = V_0/e\). Energy \(U = \frac{1}{2}CV^2\), so \(U/U_0 = (V/V_0)^2 = e^{-2} \approx 0.135\).
Incorrect. The correct answer is (A) \(e^{-2}\). \(V = V_0 e^{-1}\) at \(t=RC\). \(U \propto V^2\), so fraction = \(e^{-2} \approx 0.135\).

Free Response Questions

FRQ

F1: RC Circuit Analysis

A battery \(\mathcal{E} = 12\ \text{V}\) (negligible internal resistance) is in series with \(R = 100\ \text{k}\Omega\) and an initially uncharged \(C = 50\ \mu\text{F}\).

  1. Determine the time constant \(\tau\) of the circuit.
  2. Derive \(q(t)\) for the capacitor.
  3. Calculate the current at \(t = 2.0\ \text{s}\).
  4. Find when the capacitor reaches 99% of its final charge.
  5. Sketch \(q(t)\) and \(I(t)\) on labeled axes.
Show Solution
(a): \(\tau = RC = (100\times10^3)(50\times10^{-6}) = 5.0\ \text{s}\).
(b): Loop rule: \(\mathcal{E} - IR - q/C = 0\), \(I = dq/dt\). DE: \(\mathcal{E} = R\,dq/dt + q/C\). Solution: \(q(t) = C\mathcal{E}(1 - e^{-t/RC}) = 600\ \mu\text{C}(1 - e^{-t/5.0})\).
(c): \(I(t) = (\mathcal{E}/R)e^{-t/RC} = (1.2\times10^{-4})e^{-2.0/5.0} = 1.2\times10^{-4} \times 0.670 = 8.04\times10^{-5}\ \text{A} = 80.4\ \mu\text{A}\).
(d): \(0.99 = 1 - e^{-t/\tau}\) → \(e^{-t/\tau} = 0.01\) → \(t = -\tau\ln(0.01) = 5.0 \times 4.605 = 23.0\ \text{s}\).
(e): \(q(t)\) rises asymptotically to \(600\ \mu\text{C}\). \(I(t)\) decays exponentially from \(120\ \mu\text{A}\) to 0. At \(\tau = 5\ \text{s}\): \(q = 0.632\,C\mathcal{E}\), \(I = 0.368\,I_0\).
FRQ

F2: Ammeter & Voltmeter Design

A galvanometer has internal resistance \(r = 20\ \Omega\) and full-scale deflection at \(I_g = 1\ \text{mA}\).

  1. Design an ammeter reading up to \(1\ \text{A}\) using a shunt resistor. Draw the circuit and calculate the shunt value.
  2. Design a voltmeter reading up to \(10\ \text{V}\) using a multiplier resistor. Draw the circuit and calculate the multiplier value.
  3. Why must an ideal ammeter have zero resistance and an ideal voltmeter infinite resistance?
Show Solution
(a): Shunt \(R_s\) in parallel with galvanometer. \(I_s = 1 - 0.001 = 0.999\ \text{A}\). \(V = I_g r = 0.001 \times 20 = 0.02\ \text{V}\). \(R_s = V/I_s = 0.02/0.999 = 0.0200\ \Omega\).
(b): Multiplier \(R_m\) in series: \(10 = 0.001(20 + R_m)\) → \(R_m = 10000 - 20 = 9980\ \Omega\).
(c): Ammeter in series: if \(R \neq 0\), it changes the current being measured. Voltmeter in parallel: if \(R \neq \infty\), it draws current and changes the voltage being measured.
FRQ

F3: Kirchhoff's Rules

A circuit has \(\mathcal{E}_1 = 12\ \text{V}\) (\(r_1 = 1\ \Omega\)) in the left branch, \(\mathcal{E}_2 = 6\ \text{V}\) (\(r_2 = 1\ \Omega\)) in the right branch, and \(R_1 + R_2 = 6\ \Omega\) in the middle branch connecting top and bottom junctions.

  1. Label all currents and write Kirchhoff's junction and loop equations.
  2. Solve for all three branch currents.
  3. Calculate the power delivered by each battery. Is battery 2 being charged or discharged?
Show Solution
(a): \(I_1\) up through \(r_1\), \(I_2\) up through \(r_2\), \(I_3\) down through middle. Junction: \(I_1 + I_2 = I_3\). Left loop: \(12 - I_1(1) - I_3(6) = 0\). Right loop: \(6 - I_2(1) - I_3(6) = 0\).
(b): From loops: \(I_1 = 12 - 6I_3\), \(I_2 = 6 - 6I_3\). Junction: \((12-6I_3) + (6-6I_3) = I_3\) → \(18 = 13I_3\) → \(I_3 = 1.385\ \text{A}\). \(I_1 = 3.69\ \text{A}\), \(I_2 = -2.31\ \text{A}\) (flows opposite assumed direction).
(c): \(P_1 = \mathcal{E}_1 I_1 = 12 \times 3.69 = 44.3\ \text{W}\) (delivering). \(P_2 = 6 \times 2.31 = 13.9\ \text{W}\). Current enters + terminal of battery 2 → being charged.
FRQ

F4: Bridge Circuit Analysis

Four identical resistors \(R = 12\ \Omega\) form a bridge: \(R\) between A-B, B-C, A-D, D-C, with a wire between B and D. Battery of \(24\ \text{V}\) connected from A to C.

  1. Find the equivalent resistance between A and C.
  2. Calculate the total current from the battery.
  3. Find the current in the wire connecting B and D.
  4. Determine the power dissipated in each resistor.
Show Solution
(a): By symmetry, V_B = V_D. No current in B-D wire. Each path: \(2R = 24\ \Omega\) in series. Two paths in parallel: \(R_{\text{eq}} = 24/2 = 12\ \Omega\).
(b): \(I_{\text{total}} = V/R_{\text{eq}} = 24/12 = 2.0\ \text{A}\).
(c): V_B = V_D, so \(I_{BD} = 0\).
(d): Current splits equally: 1.0 A per branch. \(P = I^2R = 1.0^2 \times 12 = 12\ \text{W}\) each. Total = 48 W (\(P_{\text{total}} = VI = 24 \times 2 = 48\ \text{W}\)).
FRQ

F5: Resistivity & Drift Velocity

A copper wire (\(\rho = 1.72 \times 10^{-8}\ \Omega\cdot\text{m}\), \(n = 8.5 \times 10^{28}\ \text{m}^{-3}\)) has \(L = 2.0\ \text{m}\), \(d = 1.0\ \text{mm}\), and carries \(I = 3.0\ \text{A}\).

  1. Calculate the resistance of the wire.
  2. Find the drift velocity of electrons in the wire.
  3. Calculate the electric field inside the wire.
  4. If the wire is cut in half and the halves are connected in parallel to the same voltage, by what factor does power change?
Show Solution
(a): \(A = \pi(0.5\times10^{-3})^2 = 7.854\times10^{-7}\ \text{m}^2\). \(R = \rho L/A = (1.72\times10^{-8})(2.0)/(7.854\times10^{-7}) = 0.0438\ \Omega\).
(b): \(v_d = I/(neA) = 3.0/[(8.5\times10^{28})(1.6\times10^{-19})(7.854\times10^{-7})] = 2.81 \times 10^{-4}\ \text{m/s}\).
(c): \(E = \rho J = \rho I/A = (1.72\times10^{-8})(3.0/7.854\times10^{-7}) = 0.0657\ \text{V/m}\). Or \(E = V/L = IR/L = 0.0657\ \text{V/m}\).
(d): Each half: \(R_{\text{half}} = R/2\). Parallel: \(R_{\text{new}} = R/4\). \(P_{\text{new}}/P_0 = (V^2/R_{\text{new}})/(V^2/R) = 4\). Power increases by factor of 4.

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