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)
A circuit consists of an ideal battery of emf Ɛ and four identical resistors \(R_{1}\) , \(R_{2}\), \(R_{3}\), and \(R_{4}\), each of resistance R, as shown in Figure 1.
(a) For parts (a)(i) and (a)(ii), express your answers in terms of numerical values, Ɛ, and R only.
i. Derive an expression for the current \(I_{1}\) in Resistor \(R_{1}\).
ii. Derive an expression for the current \(I_{3}\) in Resistor \(R_{3}\).
(b) The partially completed bar chart in Figure 2 shows a bar that represents the absolute value \(\left| \Delta V\right|\) of the potential difference across the ideal battery.
• In Figure 2, draw a bar to represent \(\left| \Delta V\right|\) across each resistor, relative to the emf Ɛ of the ideal battery.
• The height of each bar should be proportional to the value of \(\left| \Delta V\right|\) represented by that bar. If \(\left| \Delta V\right|\) is zero, write a “0” in that column
A student claims that the rate at which energy is dissipated (power) by the circuit can be expressed as \(p = \frac{3\varepsilon ^{2}}{5R}\)
(c) State whether the expression for P is correct or incorrect. Justify your answer by referring to the derivations from part (a) or the bar chart from part (b).
When the ideal battery is connected in the original circuit, the rate at which energy is dissipated by Resistor \(R_{1}\) is \(P_{original}\). The ideal battery is now replaced with a nonideal battery of emf Ɛ and internal resistance r to form the new circuit shown in Figure 3. The rate at which energy is dissipated by Resistor \(R_{1}\) in the new circuit is \(P_{new}\).
(d) Indicate whether \(P_{new}\) is greater than, less than, or equal to \(P_{original}\).
_____ \(P_{new}\)> \(P_{original}\)_____ \(P_{new}\) < \(P_{original}\) _____ \(P_{new}\) = \(P_{original}\)
Briefly justify your answer
▶️Answer/Explanation
1(a(i))Example Response
Determine the total resistance of the circuit.
The resistance of the right-most branch containing resistors connected in series:
\(R_{s} =\sum_{i}^{}R_{i}\)
\(R_{s} =R+R\)
\(R_{s} =2R\)
The resistance of parallel branches that contain resistors:
\(\frac{1}{R_{p}} = \sum_{i}^{}\frac{1}{R_{i}}\)
\(\frac{1}{R_{p}} = \frac{1}{R} + \frac{1}{2R} = \frac{3}{2R}\)
\(R_{p} =\frac{2R}{3}\)
The total resistance of the circuit:
\(R_{s} =\sum_{i}R_{i}\)
\(R_{total} =R+\frac{2R}{3} =\frac{5R}{3}\)
The total current in the circuit:
\(I =\frac{\Delta V}{R} = \frac{3\varepsilon }{5R}\)
1(a(ii))Example Response
1(b)Example Response
1(c)Example Solution
The equation is correct. The equation for P , which is power, can be written as (\(p= \frac{\left ( \Delta v \right )^{2}}{R}\). According to the bar chart in part (b), the potential difference across the battery is \(\varepsilon \). The total resistance of the circuit is \(\frac{5R}{3}\) , according to the derivation from part (a)(i). Therefore, (\(p= \frac{\left ( \Delta v \right )^{2}}{R}\) = \(\frac{\left ( \varepsilon \right )^{2}}{\left ( \frac{5R}{3} \right )}\) = \(\frac{3\varepsilon ^{2}}{5R}\)
1(d)Example Response
\(P_{new}\) < \(P_{original}\). Since the emf of the battery is the same in the new circuit and the total resistance of the new circuit is greater, the current in \(R_{1}\) is less in the new circuit. Therefore, \(P_{new}\) is less than \(P_{original}\)
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