AP Physics C E and M Unit 11 Electric Circuits

[Maximum number: 1]

26. A 9 V battery is connected to a $450 \Omega$ load. If the internal resistance of the battery is negligible, how long will it take for 2 C to pass through the load?

A

0.01 s

B

0.02 s

C

25 s

D

50 s

E

100 s

Figure for Question 11 Electric Circuits question 1 — AP Physics C: Electricity & Magnetism

[Maximum number: 13]

The plates of a certain variable capacitor have an adjustable area. An experiment is performed to study the potential difference across the capacitor as it discharges through a resistor. A circuit is to be constructed with the following available equipment: a single ideal battery of potential difference $\Delta V_{0}$ , a single voltmeter, a single resistor of resistance R, a single uncharged variable capacitor set to capacitance C, and one or more switches as needed.

Figure for Question 11 Electric Circuits question 2 — AP Physics C: Electricity & Magnetism

(a)

Using the symbols shown, draw a schematic diagram of a circuit that can charge the capacitor and may also be used to study the potential difference across the capacitor as it discharges through the resistor.

The capacitor is fully charged by the battery. At time t=0, the capacitor starts discharging through the resistor.

[ 4 ]

(b)

Show that the potential difference $\Delta V_{C}$ across the capacitor as a function of time t is $\Delta V_{C}(t)=\Delta V_{0} e^{-\frac{t}{R C}}$ as the capacitor discharges.

[ 3 ]

For using an appropriate loop equation

Example Response
$V_{C}-V_{R}=0$

1 point

For correctly substituting $\frac{q}{C}$ and IR in a loop equation consistent with the first point

Example Response
$\frac{q}{C}=I R$

1 point

For using a differential expression consistent with the loop rule in the first point that
1 point includes a substitution of $I=-\frac{d q}{d t}$ for the current

Example Response

\begin{aligned} & \frac{q}{C}=-\frac{d q}{d t} R \\ & -\frac{1}{R C} d t=\frac{1}{q} d q \\ & -\frac{1}{R C} \int_{t=0}^{t=t} d t=\int_{q=q_{0}}^{q=q} \frac{1}{q} d q \\ & \frac{-t}{R C}=\ln \left(\frac{q(t)}{q_{0}}\right) \therefore \frac{q(t)}{q_{0}}=e^{-t / R C} \therefore \frac{q(t)}{C}=\frac{q_{0}}{C} e^{-t / R C} \therefore V(t)=V_{0} e^{-t / R C} \end{aligned}

Scoring Note: The point can be earned regardless of the sign used in the substitution of $\frac{d q}{d t}$ for the current.

Total for part (b) 3 points

(c)

The experiment is performed using a resistor of $R=150 \mathrm{k} \Omega$ . Data for the potential difference $\Delta V_{C}$ across the capacitor as a function of t are recorded and a plot of $\ln \left(\frac{\Delta V_{C}}{\Delta V_{0}}\right)$ as a function of t is created on the graph below.

Figure for Question (c) — AP Physics C: Electricity & Magnetism

[ 4 ]

(i)

Draw the best-fit line for the data.

[ 1 ]

(ii)

Using the best-fit line, calculate a value for the unknown capacitance C.

[ 3 ]

(ii) For using an appropriate equation to relate the unknown capacitance to the data 1 point
Example Response

V(t)=V_{0} e^{-t / R C} \therefore \frac{V(t)}{V_{0}}=e^{-t / R C} \therefore \ln \left(\frac{V(t)}{V_{0}}\right)=-\frac{t}{R C}

For correctly calculating the slope using two points from the best-fit line 1 point
Example Response

\text { slope }=\frac{\Delta\left(\ln \left(\frac{V}{V_{0}}\right)\right)}{\Delta t}=\frac{(-0.70-(-0.40))}{(0.54 \mathrm{~s}-0.30 \mathrm{~s})}=-1.25 \mathrm{~s}^{-1}

Scoring Note: A response may earn this point regardless of the associated unit.
For correctly relating the slope to the unknown capacitance 1 point
Example Response

\text { slope }=-\frac{1}{R C} \therefore C=\frac{-1}{\text { slope } \times(R)}=\frac{-1}{\left(-1.25 \mathrm{~s}^{-1}\right) \times(150 \mathrm{k} \Omega)}=5 \mu \mathrm{~F}

(d)

The ideal battery is then replaced with a non-ideal battery with internal resistance r, and the experiment is repeated.

[ 2 ]

(i)

Would the slope of the graph in this final experiment change compared to the graph in part (c)? Yes No

Briefly justify your answer.

[ 1 ]

(ii)

Would the vertical intercept of the graph in this final experiment change compared to the graph in part (c)? Yes No

Briefly justify your answer.

Figure for Question (ii) — AP Physics C: Electricity & Magnetism

[ 1 ]

[Maximum number: 10]

In Experiment 1, students are asked to use a graph to determine the resistivity $\rho_{1}$ of a circuit

element that is connected to a variable power supply, as shown in Figure 1. The circuit element

is cylindrical and has uniform resistivity. The students have access to a voltmeter, an ammeter,

and a ruler.

Figure 1

(a)

Describe a procedure for collecting data that would allow the students to use a graph to

determine $\rho_{1}$ , including any steps necessary to reduce experimental uncertainty.

[ 2 ]

(b)

Describe how the collected data could be graphed and how that graph would be analyzed to

determine $\rho_{1}$ .

[ 2 ]

For indicating quantities that can be graphed appropriately to determine $\rho_{1}$ , consistent with the procedure from part A, such as potential difference as a function of current. — Point B1

Scoring Notes:
- Responses that include the reciprocals of the preceding example, in addition to other equivalent graphs, also earn this point.
- This point may be earned independently of the response in part A.

For a correct analysis of the graph, consistent with the first point of part B. — Point B2

Example Response
Graph the potential difference across the circuit element as a function of the current in the circuit element. Use the slope of the graph, which is $\frac{\rho_{1}l}{\pi r^{2}}$ , where l is the length of the circuit element and r is equal to the radius of the circuit element, to determine $\rho_{1}$ .

(c)

In Experiment 2, the students are asked to use a graph to determine the resistivity $\rho_{2}$ of solid,

cylindrical resistors made of the same material but of different lengths L. The cross-sectional

area of each resistor is $5.0 \times 10^{-6} \mathrm{~m}^{2}$ . The students directly measure the resistance R between the

ends of each resistor. Table 1 provides L and R for each resistor.

Table 1

[ 4 ]

(i)

Indicate two quantities, either measured quantities from Table 1 or additional calculated

quantities, that could be graphed to produce a straight line that could be used to

determine $\rho_{2}$ .

Vertical axis:

Horizontal axis:

[ 1 ]

For indicating appropriate quantities that could be plotted on the graph to determine $\rho_{2}$ , for example R as a function of L. — Point C1

Scoring Note: This point may be earned if the vertical and horizontal variables are reversed, or for other equivalent graphs.

(ii)

On the grid provided, create a graph of the quantities indicated in part C (i).

- Use Table 2 to record the measured or calculated quantities that you will plot.

- Clearly label the axes, including units as appropriate.

- Plot the points you recorded in Table 2.

[ 2 ]

(iii)

Draw a best-fit line for the data graphed in part C (ii).

[ 1 ]

(d)

Using the best-fit line that you drew in part C (iii), calculate an experimental value for $\rho_{2}$ .

[ 2 ]

D For correctly relating the slope of the best-fit line to $\rho_{2}$

Point D1

$\text { (e.g., } \frac{\Delta R}{\Delta L}=\text { slope }=\frac{\rho_{2}}{A} \text { ) }$

For a value for $\rho_{2}$ that is between $3.5 \times 10^{-4} \Omega \cdot \mathrm{~m}$ and $4.5 \times 10^{-4} \Omega \cdot \mathrm{~m}$

Point D2

Example Response

$\begin{aligned} & \text { slope }=\frac{\Delta R}{\Delta L} \\ & \frac{\Delta R}{\Delta L}=\frac{4.0 \Omega-2.0 \Omega}{0.050 \mathrm{~m}-0.024 \mathrm{~m}} \\ & \frac{\Delta R}{\Delta L}=77 \frac{\Omega}{\mathrm{~m}} \\ & R=\frac{\rho L}{A} \\ & R=\left(\frac{\rho_{2}}{A}\right) L \\ & \text { slope }=\frac{\rho_{2}}{A} \\ & \rho_{2}=A(\mathrm{slope}) \\ & \rho_{2}=\left(5.0 \times 10^{-6} \mathrm{~m}^{2}\right)\left(77 \frac{\Omega}{\mathrm{~m}}\right) \\ & \rho_{2} \approx 3.9 \times 10^{-4} \Omega \cdot \mathrm{~m} \end{aligned}$

[Maximum number: 5]

A rotating, circular, conducting loop of area A and resistance R is in an external uniform

magnetic field of magnitude B that is directed in the -z-direction. At time t=0, the magnetic

field is perpendicular to the plane of the loop, as shown in Figure 1. The loop is rotating with

constant angular speed $\omega$ and period T about the dashed line that is along the diameter of the

loop. The value of the magnetic flux through the loop as a function of time t is $\Phi=B A \cos (\omega t)$ .

Figure 1

(a)

On the axes shown in Figure 3, sketch a graph of the instantaneous power P dissipated by

the loop as a function of t during the time interval $0 \leq t \leq T$ .

Figure 3

[ 3 ]

For a curve that starts at the origin and has equal maximum values and equal minimum values
Scoring Note: A curve that starts and ends at P=0 and has a single maximum can earn this point.

(b)

Indicate whether the sketch you drew in part C is or is not consistent with the bars that

you drew in part A. Briefly justify your answer by referencing the functional dependence

between P and $|\varepsilon|$ .

[ 2 ]

\multirow[t]{3}{*}{D} & For correctly indicating whether the representations are consistent in parts A and C & Point D1 \\ \hline & For a correct justification that indicates that the maxima and/or minima of the graph in part C align with the bars drawn in part A because P is proportional to $\mathcal{E}^{2}$ & Point D2 \\ \hline &

Example Response
Yes, part C is consistent with part A. P is proportional to $\mathcal{E}^{2}$ . When $\|\mathcal{E}\|$ is maximum, P is maximum.

& \\ \hline \end{tabular}

AP Physics C E and M Unit 11: Electric Circuits

Covered syllabus points

Covered syllabus points

11 Electric Circuits question 1

11 Electric Circuits question 2

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

Question (i)

Question (ii)

11 Electric Circuits question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question (d)

11 Electric Circuits question 4

Question (a)

Question (b)