College Physics Quiz: Circuits With Resistors And Inductors

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Circuits With Resistors And InductorsQuestion 1 of 8

An LR circuit consists of a $0.8 \, \text{H}$ inductor in series with a $40 \, \Omega$ resistor. A $24 \, \text{V}$ DC source is connected at $t = 0$ . At what time does the power delivered to the resistor reach its maximum value?

t = 0

t = \tau/2

t = \tau

t = 2\tau

t \to \infty

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College Physics Quiz

College Physics Quiz: Circuits With Resistors And Inductors

Practice Circuits With Resistors And Inductors 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 Circuits With Resistors And Inductors, 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

At $t = 0$
At $t = \tau/2$
At $t = \tau$
At $t = 2\tau$
As $t \to \infty$ (correct answer)

Explanation: When analyzing LR circuits, you need to understand how current builds up over time and how this affects power distribution between components. In an LR circuit, when the DC source is first connected at

t = 0

, the current starts at zero and grows exponentially according to

i(t) = \frac{V}{R}(1 - e^{-t/\tau})

, where

\tau = L/R

is the time constant. Here,

\tau = 0.8/40 = 0.02

seconds. The power delivered to the resistor is

P_R = i^2R

. To find when this reaches maximum, you need to recognize that resistor power is maximized when current reaches its steady-state value of

I_{max} = V/R = 24/40 = 0.6

A. Theoretically, this occurs at

t = \infty

, but practically, the current reaches about 99.3% of its maximum value at

t = 5\tau

. Looking at the wrong answers: (A)

t = 0

is incorrect because current starts at zero, so power is zero initially. (B)

t = \tau/2

represents when current reaches only about 39% of maximum. (C)

t = \tau

is when current reaches 63% of maximum. (D)

t = 2\tau

gives about 86% of maximum current. Since none of these represents the true maximum (which occurs as

t \to \infty

), the answer must be (E) - likely "none of the above" or "at

t = \infty

." Study tip: In LR circuits, remember that maximum resistor power occurs at steady state, not at some intermediate time constant multiple. The exponential approach means you never truly reach maximum in finite time.

Question 2

In an LR circuit, the current increases from zero to $63.2\%$ of its final value in $0.015 \, \text{s}$ . If the inductance is $0.3 \, \text{H}$ , what is the resistance?

$20 \, \Omega$ (correct answer)
$15 \, \Omega$
$30 \, \Omega$
$12 \, \Omega$
$25 \, \Omega$

Explanation: When you encounter an LR circuit problem involving current growth over time, you're dealing with exponential behavior governed by the circuit's time constant. The key insight is that 63.2% represents a special milestone: it's the percentage of final current reached after exactly one time constant (

\tau

). In an LR circuit, the time constant is

\tau = \frac{L}{R}

, where

L

is inductance and

R

is resistance. Since the current reaches 63.2% of its final value in 0.015 s, this means

\tau = 0.015 \, \text{s}

. Using the time constant formula:

0.015 = \frac{0.3}{R}

Solving for resistance:

R = \frac{0.3}{0.015} = 20 \, \Omega

This confirms answer A)

20 \, \Omega

is correct. The wrong answers likely come from calculation errors or conceptual mistakes. B)

15 \, \Omega

might result from incorrectly using 0.02 s instead of 0.015 s. C)

30 \, \Omega

could come from inverting the time constant formula (using

\tau = \frac{R}{L}

instead of

\tau = \frac{L}{R}

). D)

12 \, \Omega

might arise from using an incorrect percentage milestone or making arithmetic errors. Remember this pattern: in LR circuits, one time constant always corresponds to reaching 63.2% of the final current. When you see this percentage, immediately recognize that the given time equals

\frac{L}{R}

, making the solution straightforward algebra.

Question 3

In an LR circuit with time constant $\tau = 0.025 \, \text{s}$ , what percentage of the steady-state current is reached after $0.05 \, \text{s}$ ?

$86.5\%$ (correct answer)
$63.2\%$
$95.0\%$
$75.0\%$
$98.2\%$

Explanation: When you encounter LR circuit problems involving time constants, you're dealing with exponential current growth. The key relationship is that current builds up according to

I(t) = I_{\text{steady}}(1 - e^{-t/\tau})

, where

\tau

is the time constant. To find what percentage of steady-state current is reached after 0.05 s, you need to calculate the ratio

t/\tau = 0.05/0.025 = 2

. This means you're looking at 2 time constants. Substituting into the equation:

\frac{I(0.05)}{I_{\text{steady}}} = 1 - e^{-2} = 1 - e^{-2} = 1 - 0.135 = 0.865 = 86.5\%

Therefore, answer A (86.5%) is correct. Answer B (63.2%) represents the percentage reached after exactly one time constant (

1 - e^{-1} = 0.632

). This is a common trap for students who confuse the given time with one time constant period. Answer C (95.0%) would correspond to approximately 3 time constants (

1 - e^{-3} ≈ 0.95

), suggesting the student multiplied instead of divided when finding

t/\tau

. Answer D (75.0%) doesn't correspond to any standard time constant relationship and likely represents a calculation error or confusion with other exponential processes. Remember that in LR circuits, the magic number is

e^{-1} ≈ 0.368

. After one time constant, you reach 63.2% of steady state; after two time constants, 86.5%; after three, 95.0%. Memorizing these benchmarks will help you quickly check your work on exponential growth problems.

Question 4

An LR circuit has been operating in steady state with current $I_0$ . The voltage source is suddenly reversed (changed from $+V$ to $-V$ ). What is the current immediately after the voltage reversal?

$+I_0$ (correct answer)
$-I_0$
$0$
$+2I_0$
$-2I_0$

Explanation: When analyzing LR circuits with sudden voltage changes, the key principle is that current through an inductor cannot change instantaneously. This is because inductors oppose changes in current, and any instantaneous change would require infinite voltage across the inductor. Before the voltage reversal, the circuit was in steady state with current

I_0

. In steady state, the inductor acts like a short circuit (no voltage drop), so all the applied voltage

+V

appears across the resistor, giving

I_0 = V/R

. When the voltage source suddenly reverses to

-V

, the inductor's fundamental property prevents the current from changing instantly. Therefore, immediately after the reversal, the current must still be

+I_0

. The inductor will then gradually allow the current to change over time until it reaches the new steady-state value of

-V/R = -I_0

. Looking at the wrong answers: Choice B (

-I_0

) represents the final steady-state current after the transient period, not the immediate value. Choice C (

0

) incorrectly assumes the current instantly drops to zero, violating the inductor's property. Choice D (

+2I_0

) has no physical basis and likely comes from incorrectly adding the initial current to some perceived "jump." Remember this key rule: in any circuit with inductors, current cannot change instantaneously when conditions change suddenly. The current immediately after any sudden change equals the current immediately before the change, regardless of what happens to voltage sources or switches.

Question 5

In an LR circuit, the voltage across the inductor decreases from $12 \, \text{V}$ to $3 \, \text{V}$ in $0.03 \, \text{s}$ . What is the time constant of the circuit?

$0.0216 \, \text{s}$ (correct answer)
$0.015 \, \text{s}$
$0.030 \, \text{s}$
$0.0432 \, \text{s}$
$0.0108 \, \text{s}$

Explanation: When you encounter LR circuit problems involving exponential decay, you're dealing with the fundamental time constant behavior that governs how inductors respond to voltage changes. In an LR circuit, the voltage across the inductor follows exponential decay:

V_L(t) = V_0 e^{-t/\tau}

, where

\tau

is the time constant. Given that voltage drops from 12 V to 3 V in 0.03 s, you can set up the equation:

3 = 12 e^{-0.03/\tau}

. Dividing both sides by 12:

\frac{1}{4} = e^{-0.03/\tau}

. Taking the natural logarithm:

\ln(0.25) = -\frac{0.03}{\tau}

. Since

\ln(0.25) = -1.386

, you get:

-1.386 = -\frac{0.03}{\tau}

. Solving for

\tau

\tau = \frac{0.03}{1.386} = 0.0216 \text{ s}

. Looking at the wrong answers: B) 0.015 s would result if you incorrectly used

\ln(2)

instead of

\ln(4)

, confusing the voltage ratio. C) 0.030 s assumes the time constant equals the given time interval, ignoring the exponential relationship entirely. D) 0.0432 s appears if you made an arithmetic error, possibly doubling the correct value. The correct answer is A) 0.0216 s. Remember that in exponential decay problems, always set up the ratio correctly and use natural logarithms. The time constant represents the time for a quantity to decay to

1/e

(about 37%) of its initial value, not an arbitrary fraction.

Question 6

Two LR circuits have the same resistance $R$ but different inductances $L_1 = 0.2 \, \text{H}$ and $L_2 = 0.8 \, \text{H}$ . When identical voltage sources are connected to both circuits simultaneously, what is the ratio of currents $i_1(t)/i_2(t)$ at any time $t > 0$ ?

$\frac{1 - e^{-Rt/(0.2)}}{1 - e^{-Rt/(0.8)}}$ (correct answer)
$\frac{0.2}{0.8} = 0.25$
$\frac{0.8}{0.2} = 4$
$\frac{e^{-Rt/(0.2)}}{e^{-Rt/(0.8)}} = e^{-3Rt/(0.8)}$
$1$

Explanation: When analyzing LR circuits with different inductances, you need to understand how current builds up over time in each circuit. The current in an LR circuit follows the equation

i(t) = \frac{V}{R}(1 - e^{-Rt/L})

, where the time constant

\tau = L/R

determines how quickly the current approaches its steady-state value. For circuit 1:

i_1(t) = \frac{V}{R}(1 - e^{-Rt/0.2})

For circuit 2:

i_2(t) = \frac{V}{R}(1 - e^{-Rt/0.8})

Taking the ratio:

\frac{i_1(t)}{i_2(t)} = \frac{1 - e^{-Rt/0.2}}{1 - e^{-Rt/0.8}}

This confirms answer A is correct. Answer B (0.25) incorrectly assumes the current ratio equals the inductance ratio

L_1/L_2

. This ignores the time-dependent exponential behavior entirely. Answer C (4) makes the opposite mistake, using

L_2/L_1

. Again, this treats the problem as if currents were simply proportional to inductance values. Answer D shows only the ratio of the exponential terms,

\frac{e^{-Rt/0.2}}{e^{-Rt/0.8}}

. This represents the ratio of the "remaining" current that hasn't built up yet, not the actual currents themselves. Remember: In LR circuit problems, always write out the complete time-dependent current expressions before taking ratios. The exponential terms have different time constants, so you can't simplify the ratio by canceling terms prematurely. The mathematics must reflect the full transient behavior.

Question 7

In an LR decay circuit (inductor and resistor with no voltage source), the current decreases from $2.4 \, \text{A}$ to $0.6 \, \text{A}$ in $0.08 \, \text{s}$ . What is the time constant of this circuit?

$0.058 \, \text{s}$ (correct answer)
$0.040 \, \text{s}$
$0.080 \, \text{s}$
$0.116 \, \text{s}$
$0.020 \, \text{s}$

Explanation: When you encounter an LR decay circuit problem, you're dealing with exponential decay where current follows the equation

I(t) = I_0 e^{-t/\tau}

, where

\tau

is the time constant you need to find. To solve this, you can rearrange the exponential decay equation. Taking the natural logarithm of both sides:

\ln(I/I_0) = -t/\tau

, which gives you

\tau = -t/\ln(I/I_0)

. Substituting the given values:

I_0 = 2.4 \, \text{A}

I = 0.6 \, \text{A}

, and

t = 0.08 \, \text{s}

\tau = -\frac{0.08}{\ln(0.6/2.4)} = -\frac{0.08}{\ln(0.25)} = -\frac{0.08}{-1.386} = 0.058 \, \text{s}

This confirms answer (A) 0.058 s is correct. The wrong answers represent common calculation errors: (B) 0.040 s might result from incorrectly using

\ln(2.4/0.6)

instead of the reciprocal, or making an arithmetic mistake. (C) 0.080 s is simply the given time interval, suggesting confusion between the decay time and the time constant. (D) 0.116 s could result from using the wrong logarithm base or making sign errors in the calculation. Study tip: Remember that in exponential decay problems, the time constant

\tau

represents the time it takes for the quantity to decrease to

1/e \approx 37\%

of its initial value. Always double-check that your calculated time constant makes physical sense relative to the given time interval and decay rate.

Question 8

An LR circuit with $L = 0.5 \, \text{H}$ and $R = 25 \, \Omega$ carries a steady current of $1.2 \, \text{A}$ . When a switch opens to disconnect the voltage source, how much energy is initially dissipated per second in the resistor?

$36 \, \text{W}$ (correct answer)
$30 \, \text{W}$
$1.2 \, \text{W}$
$0.36 \, \text{W}$
$72 \, \text{W}$

Explanation: When an LR circuit suddenly loses its voltage source, the inductor acts like a temporary battery, maintaining current flow through the resistor. The key insight is that at the moment the switch opens, the current cannot change instantaneously due to the inductor's property that

i_L(t) = i_L(0^-)

. Since the circuit was carrying a steady

1.2 \, \text{A}

before the switch opened, it continues carrying

1.2 \, \text{A}

immediately after. The power dissipated in the resistor is found using

P = I^2R

P = (1.2)^2 \times 25 = 1.44 \times 25 = 36 \, \text{W}

This confirms answer A is correct. Looking at the wrong answers: B (

30 \, \text{W}

) likely comes from miscalculating

(1.2)^2 = 1.2

instead of

1.44

, then multiplying by

25

. C (

1.2 \, \text{W}

) suggests using

P = I \times R = 1.2 \times 1 = 1.2

, which incorrectly treats resistance as

1 \, \Omega

or confuses the current value with power. D (

0.36 \, \text{W}

) might result from using

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

with an incorrectly calculated voltage, or from computational errors. Remember: In LR circuits, current cannot change instantaneously when the switch opens. Always use the current value from just before the switching event to find the initial power dissipation. The current will then decay exponentially, but the question asks specifically about the initial moment.