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AP Physics C E&M - 11.6 Kirchhoff’s Loop Rule- Exam Style questions- MCQs

Question

A battery with emf \( \varepsilon \) and internal resistance of \(30~\Omega\) is being recharged by connecting it to an outlet with a potential difference of \(120~\mathrm{V}\), as shown above. While it is being recharged, \(3~\mathrm{A}\) flows through the battery. Determine the emf of the battery.

(A) \(210~\mathrm{V}\)
(B) \(150~\mathrm{V}\)
(C) \(90~\mathrm{V}\)
(D) \(30~\mathrm{V}\)
(E) \(9~\mathrm{V}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

Since the battery is being charged, the charging current flows into its positive terminal.

Apply Kirchhoff’s Loop Rule around the circuit:

\( 120-\varepsilon-Ir=0 \)

Substitute the given values:

\( 120-\varepsilon-(3)(30)=0 \)

\( 120-\varepsilon-90=0 \)

\( \varepsilon=30~\mathrm{V} \)

Therefore, the emf of the battery is

\( \boxed{30~\mathrm{V}} \)

Hence, the correct answer is (D).

Question

In the circuit shown above, what is the current through the \(3~\Omega\) resistor?

(A) \(0~\mathrm{A}\)
(B) \(0.5~\mathrm{A}\)
(C) \(1.0~\mathrm{A}\)
(D) \(1.5~\mathrm{A}\)
(E) \(2.0~\mathrm{A}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{E}} \)

Assume the current \(I\) flows clockwise around the loop.

Applying Kirchhoff’s Loop Rule, the algebraic sum of the potential changes around the loop is zero:

\( +7-I(3)+3-I(2)=0 \)

Simplifying,

\( 10-5I=0 \)

\( I=\dfrac{10}{5}=2.0~\mathrm{A} \)

Since the circuit contains only one loop, the same current flows through every component, including the \(3~\Omega\) resistor.

Therefore, the current through the \(3~\Omega\) resistor is

\( \boxed{2.0~\mathrm{A}} \)

Hence, the correct answer is (E).

Question

In the circuit above, the emfs and the resistances have the values shown. The current in the circuit is \(2~\mathrm{A}\).

What is the potential difference between points \(X\) and \(Y\)?

(A) \(1.2~\mathrm{V}\)
(B) \(6.0~\mathrm{V}\)
(C) \(8.4~\mathrm{V}\)
(D) \(10.8~\mathrm{V}\)
(E) \(12.2~\mathrm{V}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

From the previous part, the unknown resistance is

\( R=1~\Omega \)

Apply Kirchhoff’s Loop Rule from point \(X\) to point \(Y\) through the right-hand side of the circuit.

The potential changes are:

Across the \(6~\mathrm{V}\) battery: \( -6~\mathrm{V} \)

Across the \(0.2~\Omega\) resistor: \( -(2)(0.2)=-0.4~\mathrm{V} \)

Across the \(1~\Omega\) resistor: \( -(2)(1)=-2.0~\mathrm{V} \)

Thus,

\( V_Y-V_X=-6-0.4-2=-8.4~\mathrm{V} \)

Therefore, the magnitude of the potential difference between \(X\) and \(Y\) is

\( \boxed{|V_X-V_Y|=8.4~\mathrm{V}} \)

Hence, the correct answer is (C).

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