RC & RL transients: practice problems with step-by-step solutions
Also called: RC circuit · first-order circuits · capacitor charging
Free RC & RL transients practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: RC charging, RL circuit, capacitor with initial charge.
Free practice problemEasy
The capacitor is uncharged when the switch closes at t = 0. Find the capacitor voltage vC at t = 10 ms.
When you close a switch in a circuit with only resistors, every current jumps to its new value at once.
A capacitor or an inductor changes this. They store energy, and stored energy cannot change in zero time. So the circuit moves smoothly from the old state to the new one.
This smooth change is called a transient. After a while the circuit settles, and nothing changes any more.
Think of filling a bucket through a narrow hose. The bucket is the capacitor, the narrow hose is the resistor. The level rises fast at first, then slower and slower as the bucket gets full.
Self-check An uncharged capacitor is connected to a 12 V source through a resistor. What is the capacitor voltage just after the switch closes?
0 V
12 V
4.4 V
6 V
Answer: 0 V. A capacitor voltage cannot jump. It was 0 V just before the switch closed, so it is still 0 V just after. It reaches 12 V only after a long time.
2Capacitors and inductors at the start and the end
You only need two rules for each part: what cannot jump, and how it behaves once everything has settled.
Capacitor: its voltage vC cannot jump. Long after switching, no current flows into it, so it acts as an open circuit (a gap).
Inductor: its current iL cannot jump. Long after switching, it has no voltage across it, so it acts as a short circuit (a wire).
The value just after switching is the initial value, x(0). The value after settling is the final value, x(∞).
For a capacitor charging from a source Vs: vC(0) = 0 and vC(∞) = Vs. For an inductor: i(0) = 0 and i(∞) = Vs/R.
Self-check A 12 V source drives a 20 Ω resistor in series with an inductor. What current flows long after the switch closes?
0 A
0.6 A
1.67 A
240 A
Answer: 0.6 A. Once settled, the inductor is a plain wire, so only the resistor limits the current: 12/20 = 0.6 A. The 0 A is the value just after switching, not at the end.
3The time constant
How fast the change happens is set by one number, the time constant τ (tau).
τ = R·C τ = L/R
Units: kΩ times µF gives milliseconds. mH divided by Ω also gives milliseconds.
Every voltage and current follows the same exponential curve between its initial and final value:
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
After 1τ the change is 63% complete. After 2τ it is 86%, after 3τ 95%.
After about 5τ the circuit has settled for all practical purposes.
For a charging capacitor that starts at 0 V, the formula becomes:
vC(t) = Vs·(1 − e−t/τ)
Self-check What is the time constant of a 4.7 kΩ resistor with a 10 µF capacitor?
0.47 ms
2.13 ms
47 ms
47 s
Answer: 47 ms. τ = R·C = 4.7 kΩ · 10 µF = 47 ms, because kΩ times µF gives milliseconds. Dividing instead of multiplying gives 2.13 ms.
4The method step by step
Find the initial value x(0) just after switching. Use the rule: vC and iL cannot jump.
Find the final value x(∞). Replace a capacitor by a gap and an inductor by a wire, then solve the DC circuit.
Find the time constant: τ = R·C or τ = L/R, using the resistance the capacitor or inductor sees.
Put the three values into x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.
Fill in the time t, in the same unit as τ, and calculate.
If the question asks for a time instead, solve the same formula for t. You will need the natural logarithm, ln.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
The capacitor is uncharged when the switch closes at t = 0. Find the time constant and the capacitor voltage vC at t = 100 ms.
1
Look at the circuit
The switch connects Vs = 10 V to the capacitor through R = 10 kΩ. C = 10 µF.
2
Initial and final value
The capacitor is uncharged and its voltage cannot jump, so vC(0) = 0 V.
Long after, no current flows, so there is no voltage across R. All 10 V sits across C.
vC(0) = 0 V, vC(∞) = 10 V
3
Time constant
τ = R·C = 10 kΩ · 10 µF = 100 ms
4
Voltage at t = 100 ms
Here t is exactly one time constant, so t/τ = 1.
vC = 10·(1 − e−1) = 10·(1 − 0.368) = 6.32 V
5
Check the answer
After one time constant the capacitor should be 63% charged. 63% of 10 V is 6.3 V, so the answer fits.
It is also below the 10 V source, as a charging capacitor must be.
Answer
τ = 100 ms and vC = 6.32 V at t = 100 ms.
6Common mistakes
Using the charging formula Vs·(1 − e−t/τ) when the capacitor already holds a voltage. Use the general formula with x(0).
Mixing units: kΩ times µF is milliseconds, not seconds. Keep t and τ in the same unit.
Answering with Vs·e−t/τ. That is the part still missing (or the resistor voltage), not the capacitor voltage.
Using the wrong resistance in τ. Use the resistance the capacitor or inductor sees.
Self-check A capacitor holds 4 V when it is switched onto a 10 V source. What is vC after one time constant? (e−1 ≈ 0.368)
10.32 V
2.21 V
6.32 V
7.79 V
Answer: 7.79 V. vC = 10 + (4 − 10)·0.368 = 7.79 V. The charging formula alone gives 6.32 V, which ignores the starting 4 V. 2.21 V is the part still missing.
7How exam questions look
Easy: RC charging from 0 V, as in the example. You find τ or the capacitor voltage at a given time.
Medium: an RL circuit with the inductor current starting at 0. You find i(t) = (Vs/R)·(1 − e−t/τ), the inductor voltage vL = Vs·e−t/τ, or the time to reach a percentage of the final current.
Hard: a capacitor that already holds a voltage. You find vC at a given time, or the time to reach a given voltage.
To find a time, rearrange the general formula:
t = τ·ln[(x(0) − x(∞)) / (x(t) − x(∞))]
With τ = L/R, mH divided by Ω gives ms. So a 50 mH inductor with 10 Ω has τ = 5 ms.
The capacitor has been charged to 5 V. At t = 0 the switch closes and the capacitor discharges through R. Find the capacitor voltage vC at t = 50 ms.
vC(0) = 5 VR = 1 kΩC = 68 µFt = 50 ms
Find vC(V)
Show hints
A first-order RC circuit has time constant τ = RC. With R in kΩ and C in µF, τ comes out in ms.
With no source the capacitor voltage decays towards zero: vC(t) = vC(0)·e−t/τ.
τ = 68 ms.
Show the answer and the step-by-step solution
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Time constant
kΩ times µF gives milliseconds.
τ = R·C = 1 kΩ · 68 µF = 68 ms
Discharge curve
The capacitor starts at its stored voltage and decays towards 0 V.
vC(t) = 5·e−50/68 = 2.397 V
Answer
vC = 2.397 V
Example 2Medium
The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 75% of its final value? Find t.
Vs = 6 VR = 50 ΩL = 150 mH
Find t(ms)
Show hints
For an RL circuit τ = L/R. With L in mH and R in Ω, τ comes out in ms.
Just after the switch closes the inductor current is still zero; long after, the inductor acts like a short, so i(∞) = Vs/R.
τ = 3 ms, i(∞) = 0.12 A.
Show the answer and the step-by-step solution
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Time constant and final current
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
τ = L/R = 150 mH / 50 Ω = 3 ms
i(∞) = Vs/R = 6/50 = 0.12 A
Solve for the time
0.75 = 1 − e−t/τ → t = −τ·ln(1 − 0.75)
t = −3·ln(0.25) = 4.159 ms
Answer
t = 4.159 ms
Example 3Hard
Just before the switch closes at t = 0, the capacitor holds 18 V with the polarity shown. How long does it take for vC to reach 8.4 V? Find t.
Vs = 6 VR = 1 kΩC = 100 µFvC(0) = 18 V
Find t(ms)
Show hints
Use the general first-order solution x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.
Here x(0) is the stored capacitor voltage and x(∞) is the source voltage, because no current flows in the capacitor once it has settled.
τ = 100 ms.
Show the answer and the step-by-step solution
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Initial value, final value, time constant
The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.
t = τ·ln((18 − 6)/(8.4 − 6)) = 100·ln(5) = 160.9 ms
Answer
t = 160.9 ms
Questions and answers
What is RC & RL transients?
When you close a switch in a circuit with only resistors, every current jumps to its new value at once.
A capacitor or an inductor changes this. They store energy, and stored energy cannot change in zero time. So the circuit moves smoothly from the old state to the new one.
How do I solve RC & RL transients problems step by step?
Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.
What are common mistakes in RC & RL transients?
Using the charging formula for a discharging circuit.
Mixing units: kΩ times µF gives ms, not s.
Using the wrong resistance in the time constant.
What do the RC & RL transients exercises cover?
Easy: RC charging
Medium: RL circuit
Hard: capacitor with initial charge
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.