Resistor-Capacitor (RC) Circuits AP Physics 2 FRQ – Exam Style Questions etc.
Resistor-Capacitor (RC) Circuits AP Physics 2 FRQ
Unit 11: Electric Circuits
Weightage : 15–18%
Exam Style Practice Questions ,Resistor-Capacitor (RC) Circuits AP Physics 2 FRQ
Question
(12 points, suggested time 25 minutes)
Students are given an unknown circuit component that is connected in series to a resistor with known resistance 500 Ω.
(a) The students are asked to experimentally determine whether the component is a resistor or an uncharged capacitor.
i. Complete the following diagram to show how to use standard circuit equipment to determine whether the component is a resistor or an uncharged capacitor.
ii. Describe an experimental procedure to determine whether the component is a resistor or an uncharged capacitor. Refer to the circuit equipment in the diagram drawn in part (a)(i).
iii. What results would the students expect if the component is an uncharged capacitor? Support your answer in terms of potential difference and charge.
The students conduct a different experiment to determine the emf Ɛ of a battery that is not ideal and has internal resistance r = 30 Ω. The battery is connected to a variable resistor in a circuit, as shown. The students measure the current I through the circuit for different values of resistance \( R_{var}\) of the variable resistor that is connected to the battery. The following table contains the data collected.
(b)
i. Write an equation describing the circuit in terms of Ɛ, I, r, and \( R_{var}\).
ii. Which quantities could be graphed to yield a straight line that could be used to calculate a numerical value for the emf Ɛ of the battery?
Horizontal Axis:_______________ Vertical Axis:_______________
iii. Plot the data points for the quantities indicated in part (b)(ii) on the following graph. Clearly scale and label all axes, including units. Draw a straight line that best represents the data. You may use the blank columns in the table for any quantities you graph other than the given data.
iv. Using the graph from part (b)(iii), determine the emf Ɛ of the battery.
▶️Answer/Explanation
1(a)(i)Example Response
1(a)(ii) Example Responses
Measure the current in the ammeter immediately after the circuit is closed and a long time after the circuit is closed.
OR
Measure the potential difference across the circuit component immediately after the circuit is closed and a long time after the circuit is closed.
OR
Measure the potential difference across the circuit component a long time after the circuit is closed.
OR
Measure the potential difference across the 500 Ω resistor immediately after the circuit is closed and a long time after the circuit is closed.
1(a)(iii) Example Responses
Immediately after the circuit has been closed, a current should be measured. A long time after the circuit has been closed, a current of zero should be measured. This is because the initially uncharged capacitor becomes fully charged. This results in a potential difference across the capacitor that is equal to the potential difference across the battery a long time after the circuit has been closed. Therefore, according to Kirchhoff’s loop rule, a potential difference will not be measured across any other circuit components.
1(b)(i) Example Responses
\(0 =+\epsilon -Ir -IR_{var}\)
1(b)(ii) Example Responses
1(b)(iii) Example Responses
OR
1(b)(iv)Example Responses
Question
An uncharged air-filled capacitor has square plates of side L whose separation can be varied. The initial plate separation is \(d_{1}\), which is comparable to L. The capacitor is connected to a battery with potential difference \(V_{0}\) and a switch S, as shown above. Assume the dielectric constant of the air is 1.0.
(a) The switch is closed and remains closed for a long time.
i. On the diagram below, draw a vector at each dot to represent the electric field created by the charge on the capacitor plates.
ii. The plates are now moved to a new separation \(d_{2}\) << \(d_{1}\) and allowed to reach equilibrium. Calculate the magnitude of the charge on each plate of the capacitor in terms of \(V_{0}\), L, \(d_{2}\), and physical constants, as appropriate.
(b) The plates are pulled apart to a separation of 2\(d_{2}\), where 2\(d_{2}\) << \(d_{1}\).
i. Suppose the switch remains closed while the plates are pulled apart. Compare the magnitude of the new electric field between the plates when their separation is 2\(d_{2}\) to the magnitude of the field for plate separation \(d_{2}\). Support your comparison using appropriate physics principles.
ii. Suppose the switch is opened before the plates are pulled apart. Compare the magnitude of the new electric field between the plates when their separation is 2\(d_{2}\) to the magnitude of the field for plate separation \(d_{2}\). Support your claim using appropriate physics principles.
(c) The new circuit shown below is now assembled. The plates of the original capacitor, \(C_{1}\), are at separation \(d_{2}\). A second capacitor, \(C_{2}\), with square plates of side 2L and plate separation \(d_{2}\), is connected in parallel with the original capacitor. Both capacitors are initially uncharged. The switch is closed and the circuit reaches equilibrium. Let \(Q_{0}\) be the magnitude of the charge on each plate of capacitor \(C_{1}\) in the previous circuit, as determined in part (a)(ii). In terms of \(Q_{0}\), calculate the magnitude of the charge on each plate of the two capacitors in the new circuit.
Answer/Explanation
Ans:
(a) i) 
ii) \(Q=C\Delta V=(\varepsilon _{0}L^{2}/d_{2})V_{0}\)
\( Q=\varepsilon _{0}L^{2}V_{0}/d_{2}\)
(b) i) Since the battery remains connected, the potential difference across the capacitor remains the same. The electric field must decrease since \(E=\Delta V/d\) and d has increased
ii) The charge on the plates does not change, so E remains the same. \(E=\Delta V/d\) = Q/Cd. Since C \(\alpha\) 1/d , E is independent of the plate spacing which is the only thing that changes.
(c) The conditions for \(C_{1}\) are the same as in part (a)(ii), so it has a charge \(Q_{0}\). The area of \(C_{2}\) is four times that of \(C_{1}\), so it has four times the capacitance. In order to have the same potential difference as \(C_{1}\), it must therefore have four times the charge.
Question
The circuit shown is built and the voltage source supplies voltage for 15 minutes before the battery is completely drained. Assume the voltage supplied by the battery is constant at 12 V until the battery is drained, after which the battery supplies o V.
(a) What is the equivalent resistance of the circuit?
(b) Two students are discussing the apparatus. Student 1 says, “If the 20 n resistor were not present, the overall resistance of the circuit would have been lower and the battery would have lasted longer.” Student 2 says, “If the 20 n resistor were not present, I think the power output would have been higher and the battery would have drained faster.”
(i) Use equations to show whether the overall resistance would have been lower without the 20Ω resistor present.
(ii) Use equations to determine whether the overall power output would have been higher.
(iii) Which student is correct about the battery life?
(c) The 20 Ω resistor is replaced with a capacitor.
(i) As soon as the circuit is connected, explain without using equations how the current drawn out of the battery compares between the original circuit and the circuit with
the capacitor.
(ii) After the capacitor has been connected for a long time, but before the battery is completely drained, how does the current drawn out of the battery compare between the
original circuit and the circuit with the capacitor?
Answer/Explanation
Ans:
(a) The resistors in parallel are combined first.
Req-1 = 40-1 + 60 -1
Req = 40
The total equivalent resistance is then Rtotal = 24 + 20 = 44 Ω.
(b) (i) Without the resistor, there would only be the parallel resistor combination.
R-1 total = 40-1 + 60 -1
R total = 40 Ω.
(ii) P =IV= \(\left ( \frac{V}{R} \right )V,\) so a higher resistance results in a lower power output for the same battecy.
(iii) A lower power output means that it will take more time for the energy to be used up, so student 1 is correct.
(c) (i) As soon as the circuit is connected with the capacitor, the capacitor acts as if it had no resistance. This is exactly like
(b, i). The lower resistance would cause a greater current to be drawn from the battecy.
(ii) After the capacitor is fully charged, no more current can flow in the branch of the circuit with the capacitor. The capacitor is in series with the battecy, so there will be no
current drawn from the battecy.