College Physics Quiz: Electric Power

8 questions · exam conditions

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Electric PowerQuestion 1 of 8

A 12 V battery is connected to a circuit containing three resistors: $R_1 = 4.0 \, \Omega$ , $R_2 = 6.0 \, \Omega$ , and $R_3 = 12 \, \Omega$ . $R_1$ and $R_2$ are connected in parallel, and this combination is in series with $R_3$ . What is the power dissipated by $R_2$ ?

2.4 W

4.8 W

6.0 W

9.6 W

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College Physics Quiz

College Physics Quiz: Electric Power

Practice Electric Power in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electric Power, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

2.4 W (correct answer)
4.8 W
6.0 W
9.6 W

Explanation: First, find the equivalent resistance of

R_1

and

R_2

in parallel:

\frac{1}{R_{12}} = \frac{1}{4.0} + \frac{1}{6.0} = \frac{5}{12}

, so

R_{12} = 2.4 \, \Omega

. The total resistance is

R_{total} = 2.4 + 12 = 14.4 \, \Omega

. The total current is

I = \frac{12}{14.4} = \frac{5}{6} \, \text{A}

. The voltage across the parallel combination is

V_{12} = I \times R_{12} = \frac{5}{6} \times 2.4 = 2.0 \, \text{V}

. The current through

R_2

I_2 = \frac{2.0}{6.0} = \frac{1}{3} \, \text{A}

. The power dissipated by

R_2

P_2 = I_2^2 R_2 = (\frac{1}{3})^2 \times 6.0 = 2.4 \, \text{W}

. Choice B uses the wrong current calculation, C assumes equal current distribution, and D incorrectly applies the total voltage to

R_2

Question 2

In a DC circuit, a 60 W light bulb and a 40 W light bulb are connected in series to a battery. Both bulbs are rated for 120 V operation. Which statement correctly describes the power dissipation when connected in series?

The 60 W bulb dissipates more power because it has lower resistance and carries more current in series.
The 40 W bulb dissipates more power because it has higher resistance and the same current flows through both bulbs. (correct answer)
Both bulbs dissipate equal power because they carry the same current in a series circuit.
The total power dissipated equals 100 W because power is conserved in series circuits.

Explanation: The bulbs' ratings give us their resistances:

R_{60} = \frac{V^2}{P} = \frac{120^2}{60} = 240 \, \Omega

and

R_{40} = \frac{120^2}{40} = 360 \, \Omega

. In series, the same current flows through both, but power dissipated is

P = I^2R

. Since the 40 W bulb has higher resistance, it dissipates more power when the same current flows through both. Choice A incorrectly states current differs in series. Choice C ignores the resistance difference. Choice D incorrectly assumes the rated powers add up in series configuration.

Question 3

A battery with internal resistance $r = 2.0 \, \Omega$ and EMF $\mathcal{E} = 12 \, \text{V}$ is connected to an external resistor $R$ . For what value of $R$ will the power delivered to the external resistor be maximized?

$R = 1.0 \, \Omega$ because this minimizes the total resistance and maximizes current flow.
$R = 2.0 \, \Omega$ because maximum power transfer occurs when load resistance equals source resistance. (correct answer)
$R = 4.0 \, \Omega$ because this creates the optimal voltage division for maximum power transfer.
$R = 6.0 \, \Omega$ because this minimizes power loss in the internal resistance while maintaining sufficient current.

Explanation: The maximum power transfer theorem states that maximum power is delivered to the load when the load resistance equals the source (internal) resistance. The power delivered to the external resistor is

P = I^2 R = \left(\frac{\mathcal{E}}{R + r}\right)^2 R = \frac{\mathcal{E}^2 R}{(R + r)^2}

. Taking the derivative and setting it to zero shows that maximum occurs when

R = r = 2.0 \, \Omega

. Choice A would maximize current but not power to the load. Choice C doubles the optimal value. Choice D is too large and would reduce the power transfer.

Question 4

A 12 V car battery with negligible internal resistance is connected to a circuit containing a 3 Ω resistor in parallel with a series combination of a 4 Ω resistor and a 2 Ω resistor. What is the total power supplied by the battery?

48 W
144 W
96 W
72 W (correct answer)

Explanation: When you encounter a circuit with mixed series and parallel combinations, you need to find the equivalent resistance first, then use power formulas. Start by identifying the circuit structure: here you have a 3 Ω resistor in parallel with a series combination of 4 Ω and 2 Ω resistors. First, find the equivalent resistance of the series branch:

R_{series} = 4 + 2 = 6

Ω. Now you have 3 Ω in parallel with 6 Ω. For parallel resistors:

\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

Therefore,

R_{eq} = 2

Ω. With the total resistance known, calculate the total current:

I = \frac{V}{R} = \frac{12}{2} = 6

A. The total power supplied by the battery is:

P = VI = 12 \times 6 = 72

W. Looking at the wrong answers: Choice A (48 W) likely comes from incorrectly adding all resistances in series (3 + 4 + 2 = 9 Ω), giving

P = \frac{12^2}{9} = 16

W, or making other calculation errors. Choice B (144 W) might result from using

P = \frac{V^2}{R}

with the wrong resistance value or doubling the correct answer. Choice C (96 W) could come from using an incorrect equivalent resistance of 1.5 Ω. Study tip: Always sketch complex circuits and solve them step-by-step: series combinations first, then parallel combinations. Double-check your equivalent resistance calculation since all subsequent work depends on it.

Question 5

A student measures the power dissipated in a resistor as the applied voltage is varied. The data shows that when the voltage is doubled from 6 V to 12 V, the power increases from 2 W to 8 W. What can be concluded about the resistor?

The resistor is ohmic because power increases proportionally with voltage squared, confirming $P = \frac{V^2}{R}$ . (correct answer)
The resistor is non-ohmic because an ideal resistor would show power increasing by exactly four times when voltage doubles.
The resistor has internal heating effects that cause its resistance to increase with applied voltage.
The measurement contains experimental error because power should increase linearly with voltage for a constant resistor.

Explanation: For an ohmic resistor,

P = \frac{V^2}{R}

, so power should increase as the square of voltage. When voltage doubles, power should quadruple. Here:

\frac{8 \, \text{W}}{2 \, \text{W}} = 4

and

\frac{(12 \, \text{V})^2}{(6 \, \text{V})^2} = 4

. The data confirms

V^2

relationship, indicating ohmic behavior. We can verify:

R = \frac{V^2}{P} = \frac{36}{2} = 18 \, \Omega

at 6V, and

R = \frac{144}{8} = 18 \, \Omega

at 12V. Choice B misunderstands that the 4× increase IS the expected behavior. Choice C incorrectly interprets the correct relationship as evidence of changing resistance. Choice D incorrectly expects linear relationship.

Question 6

Two identical resistors are connected first in series, then in parallel, across the same voltage source. If the power dissipated by each resistor in the series configuration is $P_s$ , what is the power dissipated by each resistor in the parallel configuration?

$P_s$
$2P_s$
$4P_s$ (correct answer)
$8P_s$

Explanation: Let each resistor have resistance

R

and the voltage source be

V

. In series: total resistance is

2R

, current is

I_s = \frac{V}{2R}

, and power per resistor is

P_s = I_s^2 R = \frac{V^2}{4R}

. In parallel: voltage across each resistor is

V

, current through each is

I_p = \frac{V}{R}

, and power per resistor is

P_p = I_p^2 R = \frac{V^2}{R} = 4 \times \frac{V^2}{4R} = 4P_s

. Choice A assumes equal power in both configurations. Choice B accounts for voltage difference but misses the current factor. Choice D incorrectly squares the factor of 4.

Question 7

An electric kettle has two heating elements: a 1000 W element and a 1500 W element, both rated for 120 V operation. When both elements are connected in parallel to a 120 V source, what is the total power consumption?

1250 W
750 W
600 W
2500 W (correct answer)

Explanation: When you encounter questions about electrical power and parallel circuits, remember that parallel connections allow each component to operate independently at full voltage, while their currents add together. In this problem, both heating elements operate at their rated voltage of 120 V when connected in parallel to the 120 V source. This means each element consumes its full rated power: the 1000 W element draws 1000 W, and the 1500 W element draws 1500 W. Since power consumption in parallel circuits is additive, the total power is

1000 \text{ W} + 1500 \text{ W} = 2500 \text{ W}

. This confirms answer D. Answer A (1250 W) represents the average of the two power ratings, which incorrectly assumes some kind of power sharing rather than independent operation. Answer B (750 W) is the difference between the power ratings (1500 - 750 = 750), which has no physical meaning in this context. Answer C (600 W) appears to come from incorrectly treating this as a series circuit problem, where you might calculate equivalent resistance and then find power using

P = V^2/R

, but this approach is fundamentally wrong for parallel circuits. The key insight is that parallel circuits allow each device to operate at full capacity simultaneously. When you see "parallel connection" with power ratings, simply add the individual power consumptions. This is why household appliances can all operate at full power when plugged into different outlets on the same circuit.

Question 8

A space heater is designed to operate at 1500 W when connected to a 120 V outlet. Due to voltage fluctuations, the actual voltage drops to 110 V. Assuming the heater's resistance remains constant, what is the actual power consumption?

1430 W
1375 W
1265 W (correct answer)
1500 W

Explanation: When you encounter problems involving electrical devices and voltage changes, you need to think about the relationship between power, voltage, and resistance. The key insight is determining what stays constant when conditions change. Since the heater's resistance remains constant when voltage drops, you can use the power formula

P = \frac{V^2}{R}

to solve this systematically. First, find the heater's resistance using the original conditions:

R = \frac{V^2}{P} = \frac{(120)^2}{1500} = \frac{14400}{1500} = 9.6 \text{ Ω}

Now calculate the actual power at 110 V:

P = \frac{V^2}{R} = \frac{(110)^2}{9.6} = \frac{12100}{9.6} = 1260.4 \text{ W}

This rounds to approximately 1265 W, confirming answer C. Looking at the wrong answers: A) 1430 W represents a common error where students might assume power decreases linearly with voltage (110/120 × 1500 = 1375 W, then subtract some arbitrary amount). B) 1375 W comes from incorrectly using

P = VI

and assuming current stays constant, leading to the linear relationship

\frac{110}{120} × 1500 = 1375 \text{ W}

. D) 1500 W ignores the voltage change entirely. Remember this key strategy: when voltage changes but resistance stays constant in resistive devices, power changes with the square of the voltage ratio. The relationship

P = \frac{V^2}{R}

is crucial for heaters, incandescent bulbs, and other purely resistive loads. Always identify what stays constant first, then choose the appropriate power formula.