Physics Practice Test 2: Electric Circuits
- Electric Circuits
- 10 questions
- 15 minutes
This second circuits set works with combined circuits and real batteries. Simplify a circuit step by step: replace parallel groups with one equivalent resistor, then add series parts. Every answer is explained.
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Question 1
A 4.0 Ω resistor is connected in series with a parallel pair of 6.0 Ω and 12 Ω resistors. What is the equivalent resistance of the whole combination?
Correct answer: B
The parallel pair gives (6.0 × 12)/(6.0 + 12) = 4.0 Ω, and adding the 4.0 Ω in series gives 8.0 Ω. 22 Ω treats all three resistors as if they were in series.
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Question 2
The combination from the previous question (4.0 Ω in series with 6.0 Ω ∥ 12 Ω) is connected to a 24 V battery. What current flows through the 6.0 Ω resistor?
Correct answer: C
The total current is 24/8.0 = 3.0 A, so the 4.0 Ω resistor takes 12 V and 12 V is left across the parallel pair. Then I = 12/6.0 = 2.0 A. 4.0 A wrongly puts the full 24 V across the 6.0 Ω resistor.
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Question 3
A lamp is connected to an ideal battery. A second identical lamp is then connected in parallel with the first. What happens to the brightness of the first lamp?
Correct answer: A
In parallel, each lamp has the full battery voltage across it, so the current through the first lamp does not change. The battery simply supplies twice the total current; adding the lamp in series would make it dimmer.
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Question 4
A battery has an emf of 9.0 V and an internal resistance of 0.50 Ω. It is connected to a 4.0 Ω resistor. What is the voltage across the battery's terminals?
Correct answer: D
The current is 9.0/(4.0 + 0.50) = 2.0 A, so the terminal voltage is 9.0 − 2.0 × 0.50 = 8.0 V. 9.0 V ignores the voltage drop inside the battery, and 1.0 V is that drop itself.
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Question 5
Resistors of 2.0 Ω and 6.0 Ω are connected in series to a battery. How does the power dissipated in the 6.0 Ω resistor compare with that in the 2.0 Ω resistor?
Correct answer: D
In series the current is the same, so P = I²R is proportional to R: 6.0/2.0 = 3. In a parallel connection the smaller resistor would dissipate more, which is the reverse case.
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Question 6
At a junction in a circuit, 5.0 A flows in through one wire. Two wires carry current away: one carries 2.0 A. What current does the other wire carry away?
Correct answer: C
Kirchhoff's junction rule: current in equals current out, so 5.0 = 2.0 + I and I = 3.0 A. Charge is conserved, so no current can appear or disappear at the junction.
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Question 7
How should an ammeter be connected to measure the current through a resistor?
Correct answer: A
The current must flow through the ammeter, so it goes in series, and its resistance must be tiny so it does not change the current. An ammeter in parallel would act almost like a short circuit.
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Question 8
A phone battery is rated at 2000 mAh. How much charge can it deliver?
Correct answer: B
2000 mAh = 2.0 A × 3600 s = 7200 C. 120 C converts hours to minutes instead of seconds.
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Question 9
A 10 μF capacitor is connected to a 12 V battery and fully charged. How much charge is stored on each plate?
Correct answer: C
Q = CV = 10 μF × 12 V = 120 μC. 720 μC comes from ½CV², which is the stored energy in microjoules, not the charge.
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Question 10
A copper wire is 2.0 m long and has a cross-sectional area of 1.0 × 10−6 m². The resistivity of copper is 1.7 × 10−8 Ω·m. What is the wire's resistance?
Correct answer: A
R = ρL/A = (1.7 × 10−8)(2.0)/(1.0 × 10−6) = 0.034 Ω. 0.017 Ω forgets to multiply by the 2.0 m length.
Questions you leave blank are not counted as wrong.
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