AP® Physics C: E&M - Circuits with Resistors and Inductors (LR Circuits) - FRQs
Question
2. 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 Ɛ, 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.
(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 decreases with time.
ii. Describe a procedure for collecting data that would allow the students to graphically determine the experimental value for R using the measured quantity that decreases 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, Ɛ, L, t, and physical constants, as appropriate.
After reaching steady state, the absolute value of the potential difference across the inductor is \(\left| \Delta V_{1}\right|\). 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 the new inductor is \(\left| \Delta V_{2}\right|\).
(d) Indicate whether \(\left| \Delta V_{2}\right|\) is greater than, less than, or equal to \(\left| \Delta V_{1}\right|\).
_____ \(\left| \Delta V_{2}\right|\) > \(\left| \Delta V_{1}\right|\) _____ \(\left| \Delta V_{2}\right|\) < \(\left| \Delta V_{1}\right|\) _____ \(\left| \Delta V_{2}\right|\) = \(\left| \Delta V_{1}\right|\)
Justify your answer.
▶️Answer/Explanation
2(a)(i) Example Response
2(a)(ii) Example Response
Close the switch. Using the voltmeter, record the potential difference as a function of time until steady-state conditions are established.
2(b)(i) Example Response
2(b)(ii) Example Response
The data in the graph should be fit with an exponential function for the equation
\(V_{L} =\varepsilon \left ( e^{-t\frac{R}{2L}} \right )\) . Because \(\varepsilon \) and L are known, R can be calculated.
Alternate Example Response
The potential difference at 0.37\(\varepsilon \) along the vertical axis corresponds to the time constant along the horizontal axis. Because the time constant is equal to \(\frac{2L}{R}\) , and L is known, R can be calculated.
2(c) Example Solution
2(d) Example Response
\(\left| \Delta V_{2}\right|\) > \(\left| \Delta V_{1}\right|\) . The potential difference \(\left| \Delta V_{1}\right|\) across the original, ideal inductor is zero at steady state. The potential difference across the new inductor would be nonzero due to the inductor’s nonnegligible resistance. The value of \(\left| \Delta V_{2}\right|\) would be the product of the steady-state current and the inductor’s resistance.
Question
2. 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 Ɛ, 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 the 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, Ɛ, L, t, and physical constants, as appropriate.
After reaching steady state, the absolute value of the potential difference across \( R_{A}\) is \(\left| \Delta V_{1}\right|\). 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 \(\left| \Delta V_{2}\right|\).
(d) Indicate whether \(\left| \Delta V_{2}\right|\) is greater than, less than, or equal to \(\left| \Delta V_{1}\right|\).
_____ \(\left| \Delta V_{2}\right|\) > \(\left| \Delta V_{1}\right|\) _____ \(\left| \Delta V_{2}\right|\) < \(\left| \Delta V_{1}\right|\) _____ \(\left| \Delta V_{2}\right|\) = \(\left| \Delta V_{1}\right|\)
Justify your answer.
▶️Answer/Explanation
2(a)(i) Example Response
2(a)(ii) Example Response
Close the switch. Using the voltmeter, record the potential difference as a function of time until steady-state conditions are established.
2(b)(i) Example Response
2(b)(ii) Example Response
The data in the graph should be fit with an exponential function for the equation
\(V_{L} =\varepsilon \left ( e^{-t\frac{R}{2L}} \right )\) . Because \(\varepsilon \) and L are known, R can be calculated.
Alternate Example Response
The potential difference at 0.37\(\varepsilon \) along the vertical axis corresponds to the time constant along the horizontal axis. Because the time constant is equal to \(\frac{2L}{R}\) , and L is known, R can be calculated.
2(c) Example Solution
\(\varepsilon -\Delta V^{R} -\Delta V_{L} = 0\)
\(\varepsilon -I\left ( 2R \right )-L\frac{dI}{dt} = 0\)
\(\frac{\varepsilon }{L} -\frac{2IR}{L} =\frac{dI}{dt}\)
\(-\frac{R}{L}\left ( 2I – \frac{\varepsilon }{R}\right ) =\frac{dI}{dt}\)
2(d) Example Response
\(\left| \Delta V_{2}\right|\) < \(\left| \Delta V_{1}\right|\), because the inductor has nonnegligible resistance, and the total resistance of the new circuit increases as compared to the original circuit. Therefore, the current in the new circuit is reduced compared to that of the original circuit when steady-state conditions are established.
OR
\(\left| \Delta V_{2}\right|\) < \(\left| \Delta V_{1}\right|\), because the inductor has nonnegligible resistance, and the total resistance of the new circuit increases as compared to the original circuit. Therefore, the potential difference across the inductor increases, which decreases the potential difference across \( R_{A}\) compared to that of the original circuit when steady-state conditions are established.
Question
The circuit shown above is constructed using an ideal 12 V battery, an ideal switch S, and two resistors and an inductor with the values shown. Switch S is closed. After a long time, the circuit reaches steady-state conditions.
(a) Calculate the current through \(R_1\) .
(b) Calculate the current through the battery.
The switch is then opened at time t = 0.
(c) Determine the current in the inductor immediately after the switch is opened.
(d) i. Determine the current in resistor\( R_1\) immediately after the switch is opened.
ii. Which of the following statements is correct about the current through \(R_1\) immediately after the switch
is opened? _____ The current is up through \(R_1\) . _____ The current is down through \(R_1 \). _____ There is no current through \(R_1 \). Justify your answer.
(e) Immediately after the switch is opened, is the top end or bottom end of the inductor at the higher electric potential? ____ Top end ____ Bottom end Justify your answer.
(f) On the axes below, sketch a graph of the potential difference V across the inductor as a function of time after the switch is opened. Explicitly label the vertical axis intercept with a numerical value.
(g) Write but DO NOT solve a differential equation that could be solved for the current through the inductor as a function of time after the switch is opened.
Answer/Explanation
(a) For using Ohm’s law to calculate the current\( I=\frac{V_1}{R_{1}}=\frac{12V}{30\Omega }\) For a correct answer I = 0.40 A
(b) For correctly calculating the equivalent resistance of the circuit \(\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{30\Omega }+\frac{1}{20\Omega }R_{T}=12\Omega\)
For using Ohm’s law to calculate the current \(I=\frac{V}{R_{T}}=\frac{12V}{12\Omega }=1.0A\)
(c) For an answer consistent with parts (a) and (b)\( I=I_2=\frac{12V}{20\Omega }=0.60A\)
(d) i. For an answer consistent with part (c)
\( I_1 I_2 = 0.60 A\) Note: Credit is earned even if no work is shown
ii. Selecting “The current is up through R1 .” 1 point For a correct justification including inductors resisting changes in current Example Justification: Conventional current would be going down through the inductor before the switch is opened. Inductors resists changes in current, so when the switch is opened, the current would continue to be down through the inductor and up
through\( R_1\) . Note: If wrong selection is made, the justification is ignored.
(e) Select “Bottom end” or an answer consistent with part (d)(ii) For a relating the direction of the current/emf to the end of the inductor that is at the higher electric potential Example Justification: The inductor will maintain a current going down through the inductor. Since conventional current comes out of the higher potential side, the bottom end of the inductor must be at the higher potential.
(f) For a concave up curve in the first quadrant For a curve that has an asymptote at the horizontal axis For correctly indicating the maximum value
(g) For correctly applying Kirchoff’s loop rule to the circuit \(V_L-V_{R2}-V_{R1}=0\) For correctly substituting into the above equation For writing as a differential equation\( V_L-V_{R2}-V_{R1}\)
\(-L\frac{dl}{dt}=I(R_{1+R_{2}}) or (50\times 10^{-3H})\frac{dl}{dt}=I(50\Omega )\)
Question
The circuit shown above is constructed with two 6.0 V batteries and three resistors with the values shown. The currents I1 , I2 , and I3 in each branch of the circuit are indicated.
(a)
i. Using Kirchhoff’s rules, write, but DO NOT SOLVE, equations that can be used to solve for the current in each resistor.
ii. Calculate the current in the 200 Ω resistor.
iii. Calculate the power dissipated by the 200 Ω resistor.
The two 6.0 V batteries are replaced with a battery with voltage ε and a resistor of resistance 50 Ω, as shown above. The voltmeter V shows that the voltage across the 200 Ω resistor is 4.4 V.
(b) Calculate the current through the 50 Ω resistor.
(c) Calculate the voltage e of the battery.
(d)
i. The 200 Ω resistor in the circuit in Figure 2 is replaced with a 200 Fμ capacitor, as shown on the right, and the circuit is allowed to reach steady state. Calculate the current through the 50 Ω resistor.
ii. The 200 W resistor in the circuit in Figure 2 is replaced with an ideal 50 mH inductor, as shown on the right, and the circuit is allowed to reach steady state. Is the current in the 50 W resistor greater than, less than, or equal to the current calculated in part (b) ?
____ Greater than ____ Less than ____ Equal to
Justify your answer.
Answer/Explanation
Ans:
(a) i.
Loop 1: 6 – 150 I1 – 200 I2 = 0
Loop 2: 6 – 100 I3 – 200 I2 = 0
I1 + I3 = I2
ii.
I1 = I2 – I3
– 6 – 150 I2 + 150 I3 – 200 I2 = 0
6 – 150 I3 – 200 I2 = 0
——————————————
-150I2 + 250 I3 = 0
I3 = 0.6 I2
6 – 100 (0.6I2 ) – 200 I2 = 0
I2 = 0.023 (A)
iii.
P = I2R = (0.023)2 (200) = 0.107 (w)
(b)
R25 = 100 + 50 = 150 Ω
V23 = V4 = 4.4 (V)
\(I_{2}=I_{3}=\frac{V_{23}}{R_{23}}=\frac{4.4}{150}=0.029 (A)\)
(c)
\(I_{1}=I_{2}+I_{4}=I_{2}+\frac{V_{4}}{R_{4}}=0.029+\frac{4.4}{200}\)
= 0.051 (A)
\(I_{1}=I_{1}R_{1}=0.051(150)=7.7(V)\rightarrow \varepsilon =V_{1}+V_{4}=7.7 + 4.4 = 12.1 (V)\)
(d)
i. E = 12.1 (V) Capacitor is a broken circuit 7)
∑R = 150 + 100 + 50 = 300 (Ω)
\(I_{1}=I_{2}=I_{3}=\frac{12.1}{300}= 0.04 (A)\)
ii. __√__ Greater than
Because the inductor is a short circuit at steady state, the loop is series – parallel, thus, the total resistance decreases, current increases