RC Circuit Practice Problems (Time Constant, Charging, Cutoff) with Solutions

Practise how the same R and C set both the time response (charging) and the frequency response (filtering).

Key formulas

τ = R·C

V_C = V_s(1 − exp(−(t)/(τ)))

V_C = V₀·exp(−(t)/(τ))

f_c = (1)/(2πRC)

Problems 10

3 multiple-choice and 7 numeric problems. Type numbers in the unit shown next to the box (for example 2.55); SI prefixes such as 4.7k also work. Small differences from rounding along the way count as correct (±2%). When you are done, press Submit and grade at the bottom to see the answers and step-by-step solutions.

What is the time constant τ of an RC circuit with R = 4.7 kΩ and C = 10 µF, in ms?
ms
Answer: 47 ms
Solution
1. τ = R·C, with Ω and F: 10 µF = 10 × 10^(⁻⁶) F.
2. τ = 4,700 × 10 × 10^(⁻⁶) = 0.047 s = 47 ms
3. kΩ × µF gives ms directly: 4.7 × 10 = 47 ms.
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What is the cutoff frequency of a first-order RC low-pass filter with R = 1 kΩ and C = 100 nF, in kHz?
kHz
Answer: 1.592 kHz
Solution
1. f_c = (1)/(2πRC)
2. RC = 1,000 × 100 × 10^(⁻⁹) = 10^(⁻⁴) s, so f_c = (1)/(2π × 10^(⁻⁴)) = 1,592 Hz.
3. About 1.592 kHz. Forgetting the 2π gives a wrong 10 kHz.
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A capacitor charging from 0 V reaches what percentage of the source voltage after one time constant τ?

A36.8%B50%C63.2%D86.5%
Answer: C. 63.2%
Solution
1. The charging curve is V_C = V_s(1 − exp(−(t)/(τ))).
2. At t = τ, 1 − e^(⁻¹) = 0.632, i.e. 63.2%. 36.8% is what remains after τ when discharging, and 86.5% is the value after 2τ.
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A 5 V step charges R = 4.7 kΩ, C = 10 µF (τ = 47 ms) from 0 V. What is the capacitor voltage after 94 ms, in volts?
V
Answer: 4.323 V
Solution
1. 94 ms is two time constants (2τ).
2. V_C = 5 × (1 − e^(⁻²)) = 5 × (1 − 0.1353) = 4.323 V
3. After 2τ the capacitor is at 86.5% of the final value.
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R = 10 kΩ and C = 100 nF are charged from a 5 V source. When does the capacitor reach 2.5 V (half), in ms?
ms
Answer: 0.6931 ms
Solution
1. τ = 10 kΩ × 100 nF = 1 ms
2. Solving (1)/(2) = 1 − exp(−(t)/(τ)) gives t = τ·ln 2.
3. t = 1 ms × 0.6931 = 0.6931 ms. This ln 2 is the 0.693 in the 555 timer formulas.
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Roughly how long until a capacitor is charged to 99% or more of its final value?

A1τB2τC3τD5τ
Answer: D. 5τ
Solution
1. 1 − e^(⁻ⁿ): 3τ gives 95%, 4τ 98.2%, 5τ 99.3%.
2. 5τ is the first to pass 99%, so 5τ is the usual 'fully charged' time.
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A 10 µF capacitor charged to 10 V discharges through 100 kΩ. What is its voltage after 0.5 s, in volts?
V
Answer: 6.065 V
Solution
1. τ = 100 kΩ × 10 µF = 1 s
2. The discharge curve is V_C = V₀·exp(−(t)/(τ)).
3. V_C = 10 × e^(^−0.5) = 10 × 0.6065 = 6.065 V
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A 1 V amplitude, 1 kHz sine wave goes into a first-order RC low-pass filter with a 1 kHz cutoff. What is the output amplitude, in volts?
V
Answer: 0.7071 V
Solution
1. The magnitude response is (V_out)/(V_in) = (1)/(√1 + ((f)/(f_c))²).
2. At f = f_c this is (1)/(√2) = 0.7071, so the output is 0.7071 V.
3. In power that is half, or −3 dB, which is why the cutoff is called the −3 dB point.
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To make an RC filter with a 1 kHz cutoff using R = 10 kΩ, what C do you need, in nF?
nF
Answer: 15.92 nF
Solution
1. C = (1)/(2π·R·f_c)
2. C = (1)/(2π × 10,000 × 1,000) = 15.92 nF
3. With the nearest standard 15 nF, the cutoff becomes 1,060 Hz.
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At the cutoff frequency, how does the output phase of a first-order RC low-pass filter compare with the input?

ALeads the input by 45°BLags the input by 45°CLags the input by 90°DNo phase difference
Answer: B. Lags the input by 45°
Solution
1. The low-pass phase is φ = −arctan((f)/(f_c)).
2. At f = f_c, −arctan(1) = −45°: it lags by 45°. A high-pass filter leads by 45°. 90° is the limit at very high frequency.
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