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AP Physics 2- 11.6 Kirchhoff’s Loop Rule - Exam Style questions - FRQs- New Syllabus

Kirchhoff’s Loop Rule AP  Physics 2 FRQ

Unit 11: Electric Circuits

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Question

A motor is a device that when connected to a battery converts electrical energy into mechanical energy. The motor shown above is used to lift a block of mass \( M \) at constant speed from the ground to a height \( H \) above the ground in a time interval \( \Delta t \). The motor has constant resistance and is connected in series with a resistor of resistance \( R_1 \) and a battery.
Mechanical power, the rate at which mechanical work is done on the block, increases if the potential difference \( (\text{voltage drop}) \) between the two terminals of the motor increases.
(a) Determine an expression for the mechanical power in terms of \( M \), \( H \), \( \Delta t \), and physical constants, as appropriate.
(b) Without \( M \) or \( H \) being changed, the time interval \( \Delta t \) can be decreased by adding one resistor of resistance \( R_2 \), where \( R_2 > R_1 \), to the circuit shown above. How should the resistor of resistance \( R_2 \) be added to the circuit to decrease \( \Delta t \) ?
_____ In parallel with the battery      _____ In parallel with \( R_1 \)      _____ In parallel with the motor      _____ In series with the battery, \( R_1 \), and the motor
In a clear, coherent, paragraph-length response that may also contain figures and/or equations, justify why your selection would decrease \( \Delta t \).

Most-appropriate topic codes (AP Physics 2):

• Topic \( 11.4 \) — Electric Power (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \))
• Topic \( 11.5 \) — Compound Direct Current (DC) Circuits (Part \( \mathrm{(b)} \))
• Topic \( 11.6 \) — Kirchhoff’s Loop Rule (Part \( \mathrm{(b)} \))
• Skill \( 3.B \) — Apply an appropriate law, definition, theoretical relationship, or model to make a claim (Part \( \mathrm{(b)} \))
• Skill \( 3.C \) — Justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws (Part \( \mathrm{(b)} \))
▶️ Answer/Explanation

(a)
The block is lifted at constant speed, so the motor’s mechanical work goes into increasing the block’s gravitational potential energy.

The work done on the block is

\( W = MgH \)

Mechanical power is work done per unit time, so

\( P = \dfrac{W}{\Delta t} \)

Therefore,

\( \boxed{P = \dfrac{MgH}{\Delta t}} \)

Since \( M \) and \( H \) are fixed, decreasing \( \Delta t \) means the motor must deliver a larger power.

(b)
\( \boxed{\text{In parallel with } R_1} \)

To decrease \( \Delta t \), the motor must lift the same block through the same height in less time. From part \( \mathrm{(a)} \), that means the required mechanical power must increase:

\( P = \dfrac{MgH}{\Delta t} \)

The problem states that the mechanical power of the motor increases when the potential difference across the motor increases. Therefore, to reduce \( \Delta t \), we must increase the voltage across the motor.

The motor and \( R_1 \) are originally in series, so the battery voltage is shared between them. If a resistor \( R_2 \) is added in parallel with \( R_1 \), the equivalent resistance of that branch becomes smaller than \( R_1 \) alone:

\( R_{\text{eq}} = \dfrac{R_1 R_2}{R_1 + R_2} \)

so \( R_{\text{eq}} < R_1 \).

This reduces the resistance of the non-motor part of the series circuit, which increases the total current. Because the motor is still in series with the rest of the circuit, the current through the motor also increases. Since the motor has constant resistance, its voltage drop is

\( \Delta V_{\text{motor}} = I R_{\text{motor}} \)

so a larger current means a larger potential difference across the motor.

A larger voltage across the motor means a larger mechanical power output, so the motor lifts the block faster. Therefore, the time interval \( \Delta t \) decreases.

The other options do not help in the same way: putting \( R_2 \) in series would increase total resistance and reduce current, while putting it in parallel with the motor would not be the correct way to increase the motor’s power in this setup.

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