Resistance, Resistivity, and Ohm’s Law AP Physics 2 FRQ – Exam Style Questions etc.
Resistance, Resistivity, and Ohm’s Law AP Physics 2 FRQ
Unit 11: Electric Circuits
Weightage : 15–18%
Exam Style Practice Questions ,Resistance, Resistivity, and Ohm's Law AP Physics 2 FRQ
Question: (12 points, suggested time 25 minutes)
The two circuits shown above contain an ideal variable power supply, an ohmic resistor of resistance R, an ammeter A, and two voltmeters VPS and VR . In circuit 1 the ammeter has negligible resistance, and in circuit 2 the ammeter has significant internal ohmic resistance r. The potential difference of the power supply is varied, and measurements of current and potential difference are recorded.
(a) The axes below can be used to graph the current measured by the ammeter as a function of the potential difference measured across the power supply. On the axes, do the following.
• Sketch a possible graph for circuit 1 and label it 1.
• Sketch a possible graph for circuit 2 and label it 2.
(b) Let ΔVPS be the potential difference measured by voltmeter VPS across the power supply, and let I be the
current measured by the ammeter A. For each circuit, write an equation that satisfies conservation of energy,
in terms of ΔVPS, I, R, and r, as appropriate.
Circuit 1 Circuit 2
(c) Explain how your equations in part (b) account for any differences between graphs 1 and 2 in part (a).
(d) In circuit 2, R = 40 Ω . When voltmeter VPS reads 3.0 V, voltmeter VR reads 2.5 V. Calculate the internal resistance r of the ammeter.
(e) Voltmeter VR in circuit 2 is replaced by a resistor with resistance 120 Ω to create circuit 3 shown below. Voltmeter VPS still reads 3.0 V.
i. Calculate the equivalent resistance Req of the circuit.
ii. Calculate the current in each of the resistors that are in parallel.
Answer/Explanation
Ans:
(a)
V = IR
I/V = I/R
(b)
Circuit 1 Circuit 2
ΔVPS = IR ΔVPS = Ir+IR
(c)
Circuit 1’s equation can be written as \(I = \frac{I}{R}.V\) , and circuit 2’s equation as \(I = \frac{I}{r+R}.V.\) If the equations are booked at using the y=mx+b format, circuit 2 will have a smaller m, leading to the smaller slope as seen in pant a
(d)
VPS = IR + Ir = 3.0V = 2.5V + (0.0625)r
Vr = I (40Ω) = 2.5 V \(r = \frac{0.5}{0.0625}\Rightarrow 8\)
I = 0.0625A r = 8Ω
(e)
i.
\(R_{eq}=r+\left ( \left ( \frac{1}{40}+\frac{1}{120} \right )^{-1} \right )=8+30\)
Req = 38Ω
ii.
VPS = I(Req) ⇒ \(I=\frac{V}{R}= 0.07895\)
Vr = 0.6316 V/Req = I I40Ω = 0.059A
3-0.6316=2.3684 V/R120 = I I120R = 0.0197A
Question
In the circuit shown above, the current delivered by the 9-volt battery of internal resistance 1 ohm is 3 amperes. The power dissipated in R2 is 12 watts.
a. Determine the reading of voltmeter V in the diagram.
b. Determine the resistance of R2.
c. Determine the resistance of R1.
Answer/Explanation
Ans:
a. VT = E – Ir = 6 V
b. In parallel, each resistor gets 6 V and P = V2/R gives R = 3 Ω
c. For the 3 Ω resistor we have I = V/R = 2 A leaving 1 A for the branch with R1. R = V/I = 6 Ω