Practice problems

RC filter practice problems

An RC filter sorts signals by frequency, and one formula, f_c = 1 ÷ (2π R C), sets where it starts cutting. These ten problems start with finding the cutoff and work up to the output either side of it and designing a filter that cuts a whine to a tenth. Type your answer to check it, and open the working whenever you want it.

The method

f_c = 12π R C

R = 1 over 2π f_c C or C = 1 over 2π f_c R to design for a cutoff
At the cutoff, the output is 70.7% of the input: −3 dB
Well past the cutoff, the output falls to a tenth for each ×10 in frequency: −20 dB per decade, about f_c over f for a low-pass
Near the cutoff, a low-pass passes 1 over √(1 + (f ÷ f_c)²) of the input

Use ohms, farads and hertz. A high-pass has the same cutoff: swap R and C, and take the output across the resistor.

The problems

Type your answer and tap Check. Every one has the working, step by step.

Problem 1 of 10
What is the cutoff frequency of an RC low-pass filter with 4.7 kΩ and 100 nF?
Show the working
1. Use f_c = 1 over 2π R C, with R = 4700 Ω and C = 100 nF = 10⁻⁷ F.
2. 2π × 4700 × 10⁻⁷ ≈ 0.002953, so f_c = 1 over 0.002953 ≈ 339 Hz
Answer: 339 Hz
Problem 2 of 10
An RC low-pass uses 3.3 kΩ and 10 nF. What is its cutoff, in kilohertz?
Show the working
1. f_c = 1 over 2π × 3300 × 10⁻⁸ ≈ 4823 Hz
2. 4823 Hz ≈ 4.82 kHz
Answer: 4.82 kHz
Problem 3 of 10
A 2 V peak sine wave at exactly a low-pass filter's cutoff frequency goes in. What peak voltage comes out?
Show the working
1. At the cutoff X_C = R, and the output is 1 over √2 ≈ 70.7% of the input: the −3 dB point.
2. 0.707 × 2 V ≈ 1.41 V peak
Answer: 1.41 V
Problem 4 of 10
A 1 µF coupling capacitor feeds an amplifier input of 22 kΩ. What is the cutoff of this high-pass filter?
Show the working
1. A high-pass has the same cutoff formula: f_c = 1 over 2π R C, with C = 1 µF = 10⁻⁶ F.
2. f_c = 1 over 2π × 22,000 × 10⁻⁶ ≈ 7.23 Hz, far below anything you can hear, so all the audio gets through.
Answer: 7.23 Hz
Problem 5 of 10
You want a low-pass cutoff of 150 Hz with a 470 nF capacitor. What resistor do you need, in kilohms?
Show the working
1. Rearrange the cutoff formula for R: R = 1 over 2π f_c C.
2. R = 1 over 2π × 150 × 470 × 10⁻⁹ ≈ 2258 Ω ≈ 2.26 kΩ. The nearest E12 value, 2.2 kΩ, gives 154 Hz.
Answer: 2.26 kΩ
Problem 6 of 10
You want a high-pass cutoff of 50 Hz with a 10 kΩ resistor. What capacitor do you need, in nanofarads?
Show the working
1. Rearrange the cutoff formula for C: C = 1 over 2π f_c R.
2. C = 1 over 2π × 50 × 10,000 ≈ 3.18 × 10⁻⁷ F = 318 nF. The nearest E6 value is 330 nF.
Answer: 318 nF
Problem 7 of 10
A low-pass filter cuts off at 1 kHz. A 5 V peak signal at 10 kHz goes in. About how big is the output?
Show the working
1. Well above the cutoff, a low-pass passes about f_c over f of the input.
2. 1 kHz over 10 kHz × 5 V = 0.5 V peak. The exact formula gives 0.497 V.
Answer: 0.5 V
Problem 8 of 10
A low-pass filter cuts off at 300 Hz. What is its gain at 30 kHz, in decibels?
Show the working
1. 30 kHz is 100 times 300 Hz: two decades above the cutoff.
2. A first-order filter falls 20 dB per decade, so 2 × −20 dB = −40 dB, a hundredth of the voltage.
Answer: −40 dB
Problem 9 of 10
A low-pass filter cuts off at 1 kHz. What percentage of the input voltage comes out at 2 kHz?
Show the working
1. Near the cutoff, use the full formula: V_out over V_in = 1 over √(1 + (f ÷ f_c)²).
2. f ÷ f_c = 2, so V_out over V_in = 1 over √(1 + 4) = 1 over √5 ≈ 0.447
3. 0.447 is 44.7%, about −7 dB: a little below the −6 dB that the 6 dB per doubling rule gives.
Answer: 44.7%
Problem 10 of 10
A sensor's signal stays below 200 Hz, but a 20 kHz whine rides on it. You want a low-pass with a 47 nF capacitor that cuts the whine to about a tenth (−20 dB). What resistor, in kilohms?
Show the working
1. −20 dB is one decade of roll-off, so put the cutoff a decade below 20 kHz: f_c = 2 kHz.
2. R = 1 over 2π × 2000 × 47 × 10⁻⁹ ≈ 1693 Ω ≈ 1.69 kΩ. The signal at 200 Hz sits a decade below the cutoff, so it passes almost untouched.
Answer: 1.69 kΩ

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“Low- and high-pass filters” is lesson 4 of 6 in Oscillation Coast, Electronics World 4. Free with an account. Lessons unlock in order, so you’ll do the 24 before it first.

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Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Low- and high-pass filters” is lesson 4 of 6 in Oscillation Coast, Electronics World 4. Free with an account. Lessons unlock in order, so you’ll do the 24 before it first.

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The ideas behind it

Questions people ask

Why is the cutoff at 70.7% and not 50%?

At the cutoff the capacitor's reactance equals R, so the impedance is √2 × R and the output is 1 ÷ √2 ≈ 70.7% of the input. That's half the power, since 0.707² = 0.5, which is why the cutoff is also called the −3 dB or half-power point.

Is the cutoff formula the same for a high-pass filter?

Yes. Swap R and C and take the output across the resistor: the cutoff is still 1 ÷ (2π × R × C). A low-pass passes what's below the cutoff, and a high-pass passes what's above it.

What does −20 dB per decade mean?

Past the cutoff, each time the frequency goes up ten times (a decade), a first-order RC filter's output falls to a tenth, which is −20 dB. So a decade past the cutoff lets through about 10%, and two decades about 1%.