AP Physics 2- 11.2 Simple Circuits- Exam Style questions - FRQs- New Syllabus
Simple Circuits AP Physics 2 FRQ
Unit 11: Electric Circuits
Weightage : 15–18%
Question
| \( I \) \( (\mathrm{A}) \) | \( R_{\text{var}} \) \( (\Omega) \) | ||
|---|---|---|---|
| \( 0.087 \) | \( 200 \) | ||
| \( 0.060 \) | \( 300 \) | ||
| \( 0.042 \) | \( 450 \) | ||
| \( 0.027 \) | \( 700 \) | ||
| \( 0.016 \) | \( 1200 \) |
Most-appropriate topic codes (AP Physics 2):
• Topic \( 11.5 \) — Compound Direct Current (DC) Circuits (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \))
• Topic \( 11.6 \) — Kirchhoff’s Loop Rule (Part \( \mathrm{(a)(iii)} \), Part \( \mathrm{(b)(i)} \))
▶️ Answer/Explanation
(a)(i)
A correct circuit must include a source of potential difference and a measurement device. One valid arrangement is a battery, switch, and ammeter in series with the \( 500\ \Omega \) resistor and the unknown component, with a voltmeter optionally placed across the unknown component.
(a)(ii)
Close the switch and measure either:
• the current in the circuit immediately after closing the switch and again a long time later,
or
• the potential difference across the unknown component immediately after closing the switch and again a long time later,
or
• the potential difference across the \( 500\ \Omega \) resistor immediately after closing the switch and again a long time later.
Comparing the early-time and long-time behavior tells whether the unknown component is a resistor or a capacitor.
A resistor would give steady values right away, but a capacitor would give changing values with time.
(a)(iii)
If the component is an uncharged capacitor, immediately after the circuit is closed there is current in the circuit because the capacitor begins charging.
As time passes, charge builds up on the capacitor plates, so the potential difference across the capacitor increases from \( 0 \) toward the battery voltage.
Because the capacitor’s potential difference increases, the current in the series circuit decreases with time. A long time after the switch is closed, the capacitor is fully charged, so the current becomes \( 0 \).
At that long time, the potential difference across the capacitor equals the battery’s potential difference, so the potential difference across the \( 500\ \Omega \) resistor is \( 0 \).
(b)(i)
Applying Kirchhoff’s loop rule,
\( \mathcal{E} – Ir – IR_{\text{var}} = 0 \)
or equivalently,
\( \boxed{\mathcal{E}=I(r+R_{\text{var}})} \)
(b)(ii)
One valid straight-line graph is:
Horizontal Axis: \( \dfrac{1}{r+R_{\text{var}}} \) \( (\Omega^{-1}) \)
Vertical Axis: \( I \) \( (\mathrm{A}) \)
because
\( I=\mathcal{E}\left(\dfrac{1}{r+R_{\text{var}}}\right) \)
so the slope of the graph equals \( \mathcal{E} \).
Another valid choice is a graph of \( IR_{\text{var}} \) versus \( I \), where the vertical intercept is \( \mathcal{E} \). I’ll use the first option.
(b)(iii)
First calculate \( \dfrac{1}{r+R_{\text{var}}} \) using \( r=30\ \Omega \).
| \( I \) \( (\mathrm{A}) \) | \( R_{\text{var}} \) \( (\Omega) \) | \( r+R_{\text{var}} \) \( (\Omega) \) | \( \dfrac{1}{r+R_{\text{var}}} \) \( (\Omega^{-1}) \) |
|---|---|---|---|
| \( 0.087 \) | \( 200 \) | \( 230 \) | \( 0.00435 \) |
| \( 0.060 \) | \( 300 \) | \( 330 \) | \( 0.00303 \) |
| \( 0.042 \) | \( 450 \) | \( 480 \) | \( 0.00208 \) |
| \( 0.027 \) | \( 700 \) | \( 730 \) | \( 0.00137 \) |
| \( 0.016 \) | \( 1200 \) | \( 1230 \) | \( 0.000813 \) |
A correct plot and best-fit line are shown below.
(b)(iv)
Since the graph is \( I \) versus \( \dfrac{1}{r+R_{\text{var}}} \), the slope equals \( \mathcal{E} \).
Using two convenient points on the best-fit line, for example about \( (0.002\ \Omega^{-1},\,0.040\ \mathrm{A}) \) and \( (0.004\ \Omega^{-1},\,0.080\ \mathrm{A}) \),
\( \text{slope}=\dfrac{0.080-0.040}{0.004-0.002}=\dfrac{0.040}{0.002}=20 \)
so
\( \boxed{\mathcal{E}\approx 20\ \mathrm{V}} \)
This is in the expected range from the data, since larger total resistance gives smaller current in a way that is consistent with a battery emf of about \( 20\ \mathrm{V} \).