AP Physics C- E&M- 11.6 Kirchhoff’s Loop Rule- Exam Style questions - FRQs- New Syllabus
Question
Students are asked to determine the resistance \(R\) of two identical resistors. The resistors are in parallel with each other and are connected in series to a battery of known emf \(\mathcal{E}\), an inductor of known inductance \(L\), and a switch, as shown in Figure \(1\). The students have access to a voltmeter that can measure potential difference as a function of time. The students are required to measure a quantity that decreases with time to determine \(R\).
• Label the quantities graphed on the vertical and horizontal axes.
• Sketch a line or curve that represents the expected trend of the collected data.
• Label any appropriate intercepts and/or asymptotes in terms of the quantities provided.
Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):
• Topic \(11.5\) — Compound Direct Current Circuits
• Topic \(11.6\) — Kirchhoff’s Loop Rule
• Topic \(13.5\) — Circuits with Resistors and Inductors \(\left(\text{LR Circuits}\right)\)
▶️ Answer/Explanation
(a)(i)
The voltmeter should be connected in parallel with the inductor \(L\).
This measures the potential difference across the inductor:
\( |\Delta V_L| \)
Immediately after the switch is closed, the inductor opposes the change in current, so \(|\Delta V_L|\) is large. As the current approaches steady state, \(\dfrac{dI}{dt}\to 0\), so the potential difference across an ideal inductor decreases toward \(0\).
(a)(ii)
Close the switch and use the voltmeter to record the potential difference across the inductor as a function of time. Record values of \(|\Delta V_L|\) from immediately after the switch is closed until the circuit reaches steady state. Repeat the trial several times and average the measured values at corresponding times to reduce random uncertainty.
The useful data table would contain:
\( t \quad \text{and} \quad |\Delta V_L| \)
(b)(i)
The vertical axis should be labeled:
\( |\Delta V_L|\;(\text{V}) \)
The horizontal axis should be labeled:
\( t\;(\text{s}) \)
The graph should be a decreasing exponential curve. It starts at
\( |\Delta V_L|=\mathcal{E} \)
at \(t=0\), and approaches the horizontal asymptote
\( |\Delta V_L|=0 \)
as \(t\to\infty\).
(b)(ii)
The equivalent resistance of the two identical parallel resistors is
\( R_{\text{eq}}=\dfrac{R}{2} \)
For an \(LR\) circuit,
\( \tau=\dfrac{L}{R_{\text{eq}}} \)
Therefore,
\( \tau=\dfrac{L}{R/2}=\dfrac{2L}{R} \)
The graph can be fit with an exponential function:
\( |\Delta V_L|=\mathcal{E}e^{-Rt/(2L)} \)
The coefficient of \(t\) in the exponent is
\( \dfrac{R}{2L} \)
so \(R\) can be calculated from the exponential fit. Equivalently, find the time \(\tau\) when
\( |\Delta V_L|=0.37\mathcal{E} \)
and then calculate
\( \boxed{R=\dfrac{2L}{\tau}} \)
(c)
Apply Kirchhoff’s loop rule to the circuit:
\( \mathcal{E}-\Delta V_R-\Delta V_L=0 \)
The two identical resistors are in parallel, so their equivalent resistance is
\( R_{\text{eq}}=\dfrac{R}{2} \)
The potential difference across the resistor combination is
\( \Delta V_R=I\left(\dfrac{R}{2}\right) \)
The potential difference across the inductor is
\( \Delta V_L=L\dfrac{dI}{dt} \)
Substitute these into Kirchhoff’s loop rule:
\( \mathcal{E}-I\left(\dfrac{R}{2}\right)-L\dfrac{dI}{dt}=0 \)
Rearranging gives the differential equation:
\( L\dfrac{dI}{dt}+I\left(\dfrac{R}{2}\right)=\mathcal{E} \)
or
\( \boxed{\dfrac{dI}{dt}=\dfrac{\mathcal{E}}{L}-\dfrac{R}{2L}I} \)
(d)
Correct choice:
\( \boxed{|\Delta V_2|>|\Delta V_1|} \)
For the original ideal inductor, after a long time the current is steady, so
\( \dfrac{dI}{dt}=0 \)
Therefore the potential difference across the ideal inductor is
\( |\Delta V_1|=L\dfrac{dI}{dt}=0 \)
For the new inductor with nonnegligible resistance, after a long time the current is steady, but the inductor still has an ohmic voltage drop due to its internal resistance. Thus,
\( |\Delta V_2|=I_{\text{steady}}r_{\text{inductor}} \)
which is nonzero. Hence,
\( \boxed{|\Delta V_2|>|\Delta V_1|} \)