AP Physics C- E&M- 13.4 Inductance - Exam Style questions - FRQs- New Syllabus
Question
Students are asked to determine the resistance \(R\) of identical resistors \(R_A\) and \(R_B\). The resistors are connected in series with each other, 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 increases with time to determine \(R\).
(a)
i. On the circuit diagram shown in Figure \(1\), draw the voltmeter, using the following symbol, with connections that would allow the students to correctly measure a potential difference that increases with time.
ii. Describe a procedure for collecting data that would allow the students to graphically determine the experimental value for \(R\) using a measured quantity that increases with time. Provide enough detail so that another student could replicate the experiment.
(b)
i. On the axes shown in Figure \(2\), produce a graph that represents the expected trend of the data by completing the following tasks.
• 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.
ii. Describe how the information from the graph in part (b)(i) would be used to determine the experimental value for \(R\).
(c) Starting with an appropriate application of Kirchhoff’s loop rule, derive, but do not solve, a differential equation that can be used to determine the current \(I\) in the inductor at time \(t\) after the switch is closed. Express your answer in terms of \(R\), \(\mathcal{E}\), \(L\), \(t\), and physical constants, as appropriate.
After reaching steady state, the absolute value of the potential difference across \(R_A\) is \(|\Delta V_1|\). The students replace the original inductor with a new inductor that has nonnegligible resistance. The experiment is repeated. After a long time, the absolute value of the potential difference across \(R_A\) is \(|\Delta V_2|\).
(d) Indicate whether \(|\Delta V_2|\) is greater than, less than, or equal to \(|\Delta V_1|\).
_____ \(|\Delta V_2|>|\Delta V_1|\) _____ \(|\Delta V_2|<|\Delta V_1|\) _____ \(|\Delta V_2|=|\Delta V_1|\)
Justify your answer.
Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):
• Topic \(11.5\) — Compound Direct Current Circuits (Parts \( \mathrm{a} \), \( \mathrm{b} \), \( \mathrm{d} \))
• Topic \(11.6\) — Kirchhoff’s Loop Rule (Part \( \mathrm{c} \))
• Topic \(13.4\) — Inductance (Parts \( \mathrm{a} \), \( \mathrm{b} \), \( \mathrm{c} \))
• Topic \(13.5\) — Circuits with Resistors and Inductors (LR Circuits) (Parts \( \mathrm{a} \), \( \mathrm{b} \), \( \mathrm{c} \), \( \mathrm{d} \))
▶️ Answer/Explanation
(a)(i)
Place the voltmeter in parallel with either resistor \(R_A\), resistor \(R_B\), or the series combination of \(R_A\) and \(R_B\).
This works because the current in the LR circuit starts at \(0\) and increases with time, so the potential difference across any resistor also increases with time.
(a)(ii)
Close the switch and use the voltmeter to record the potential difference across \(R_A\), \(R_B\), or the combination \(R_A+R_B\) as a function of time. Record values from immediately after the switch is closed until the reading becomes approximately constant. Repeat the trial several times and average the data or use a smooth curve fit to reduce experimental uncertainty.
If the voltmeter is across one resistor, the steady-state value should approach \(\mathcal{E}/2\). If the voltmeter is across both resistors together, the steady-state value should approach \(\mathcal{E}\).
(b)(i)
The graph should have time \(t\) on the horizontal axis and potential difference \(\Delta V\) on the vertical axis. The curve starts at \(0\), increases concave down, and approaches a horizontal asymptote.
If the voltmeter is across one resistor:
\( \Delta V_R=\dfrac{\mathcal{E}}{2}\left(1-e^{-2Rt/L}\right) \)
The horizontal asymptote is \( \Delta V_R=\dfrac{\mathcal{E}}{2} \).
If the voltmeter is across both resistors:
\( \Delta V_{AB}=\mathcal{E}\left(1-e^{-2Rt/L}\right) \)
The horizontal asymptote is \( \Delta V_{AB}=\mathcal{E} \).
(b)(ii)
Fit the graph with an exponential function. The time constant for the circuit is
\( \tau=\dfrac{L}{2R} \)
On the graph, \(\tau\) is the time when the measured potential difference reaches about \(0.63\) of its final value.
Therefore,
\( \boxed{R=\dfrac{L}{2\tau}} \)
Equivalently, if the fitted exponential has the form \(1-e^{-bt}\), then \(b=\dfrac{2R}{L}\), so
\( \boxed{R=\dfrac{bL}{2}} \)
(c)
Apply Kirchhoff’s loop rule around the circuit:
\( \mathcal{E}-\Delta V_R-\Delta V_L=0 \)
Since the two identical resistors are in series, their total resistance is
\( R_{\text{total}}=R+R=2R \)
The potential difference across the two resistors is
\( \Delta V_R=I(2R) \)
The magnitude of the potential difference across the inductor is
\( \Delta V_L=L\dfrac{dI}{dt} \)
Substitute into Kirchhoff’s loop rule:
\( \mathcal{E}-2RI-L\dfrac{dI}{dt}=0 \)
Thus, a differential equation for the current is
\( \boxed{\dfrac{dI}{dt}=\dfrac{\mathcal{E}-2RI}{L}} \)
(d)
Correct choice:
\( \boxed{|\Delta V_2|<|\Delta V_1|} \)
With the original ideal inductor, after a long time the inductor acts like a wire, so the steady-state current is
\( I_1=\dfrac{\mathcal{E}}{2R} \)
Therefore, the potential difference across \(R_A\) is
\( |\Delta V_1|=I_1R=\dfrac{\mathcal{E}}{2} \)
With the new inductor, the inductor has nonnegligible resistance, so the total circuit resistance is greater than \(2R\). The steady-state current is therefore smaller than before. Since the potential difference across \(R_A\) is \(IR\), the potential difference across \(R_A\) also decreases.
Hence,
\( \boxed{|\Delta V_2|<|\Delta V_1|} \)