AP Physics C E and M 11.5 Compound DC Circuits

[Maximum number: 14]

E\&M.2.
The circuit shown above consists of a source of variable emf $\mathcal{E}$ , an ideal ammeter A, an ideal voltmeter V, a resistor of resistance R, and a sample of wire with resistance r.

(a)

How does the current through the wire sample compare with the current through the resistor R ? It is greater through R. It is greater through the sample. It is the same through both. It depends on the resistance of the sample.
Justify your answer.

[ 2 ]

(b)

How does the potential difference across the wire sample compare with the potential difference across the resistor R ? It is greater across R. It is greater across the sample. It is the same across both. It depends on the resistance of the sample.
Justify your answer.
With the sample of wire in place, the emf of the source is set to a given value. The current through and potential difference across the resistor R are measured. This is repeated for several values of emf, and the data are recorded in the table below.

Table for Question (b) — AP Physics C: Electricity & Magnetism

[ 2 ]

(c)

Indicate below which quantities should be graphed to yield a straight line that could be used to calculate a numerical value for the resistance of the wire sample.
Horizontal axis:
Vertical axis:
You may use the remaining columns in the table above, as needed, to record any quantities that you indicated that are not given.

[ 1 ]

1 point
For indicating the correct quantities for each axis Horizontal axis: $I_{R}$ Vertical axis: $V_{r}=\boldsymbol{\varepsilon}-V_{R}$ Note: Credit is received if the axes are reversed.

(d)

On the grid below, plot the straight line data points from part (c). Clearly scale and label all axes, including units if appropriate. Draw a straight line that best represents the data.

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

[ 3 ]

(e)

Use your straight line to calculate the value of the resistance of the wire sample.

[ 2 ]

2 points
For correctly calculating slope using the best-fit straight line and not data points 1 point slope $=\frac{\left(V_{2}-V_{1}\right)}{\left(I_{2}-I_{1}\right)}=\frac{(0.30-0.10) \mathrm{V}}{(0.60-0.22) \mathrm{A}}=0.526 \mathrm{~V} / \mathrm{A}$
Note: Linear regression gives slope $=0.485 \mathrm{~V} / \mathrm{A}$ . (Student must indicate that the value comes from a linear regression from the calculator, and the equation of the line must be present.)
For correctly relating r to the slope 1 point
$V_{r}=I_{R} r$ , therefore the slope equals r
$r=0.526 \Omega$
Note: Linear regression gives $r=0.485 \Omega$ .

(f)

Suppose the ammeter used to collect these data was not ideal. Would the actual value of the resistance of the wire sample be greater than, less than, or equal to that calculated in part (e) ?

Greater than Less than Equal to
Justify your answer.

If the ideal voltmeter is replaced by a voltmeter that is not ideal and the experiment is repeated, would the readings of the ideal ammeter be greater than, less than, or equal to those in the data chart before part (c) ? Greater than Less than Equal to
Justify your answer.

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

[ 4 ]

[Maximum number: 4]

Students are asked to determine the resistance R of identical resistors $\mathrm{R}_{\mathrm{A}}$ and $\mathrm{R}_{\mathrm{B}}$ . 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)

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.

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

Voltmeter Symbol

[ 1 ]

(b)

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| \quad-\left|\Delta V_{2}\right|<\left|\Delta V_{1}\right| \quad-\left|\Delta V_{2}\right|=\left|\Delta V_{1}\right|

Justify your answer.

Figure 1

[ 3 ]

For selecting $\left|\Delta V_{2}\right|<\left|\Delta V_{1}\right|$ with an attempt at a relevant justification For indicating that the total resistance has increased For indicating that the current decreases due to the increased resistance 1 point

OR

For indicating that the potential difference across the inductor will increase, and, therefore, the potential difference across $\mathrm{R}_{\mathrm{A}}$ must decrease

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 $\mathrm{R}_{\mathrm{A}}$ compared to that of the original circuit when steady-state conditions are established.

Total for part (d)

for question 2

15 points

AP Physics C E and M 11.5: Compound DC Circuits

Covered syllabus points

Covered syllabus points

11.5 Compound Direct Current Circuits question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

11.5 Compound Direct Current Circuits question 2

Question (a)

Question (b)