Practice problems

AC voltage and transformer practice problems

AC swings up and down, so it takes a few numbers to describe: peak, RMS, period and frequency. These ten problems start with reading a wave and work up to the power it delivers and what a transformer does to it. Type your answer to check it, and open the working whenever you want it.

The method

f = 1T

V_pp = 2 × V_peak: peak-to-peak is top to bottom
V_rms = V_peak over √2 ≈ 0.707 × V_peak for a sine wave
P = V_rms² over R: RMS values work in the power formulas just as DC does
V_out over V_in = N_out over N_in for a transformer, by its turns

Always say which amplitude you mean: "12 V peak" and "12 V RMS" are different waves. Mains is quoted in RMS, and its peaks can kill: these problems are sums, not something to build.

The problems

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

Problem 1 of 10
A scope shows a sine wave that's 6 V peak-to-peak. What is its peak voltage?
Show the working
1. Peak-to-peak runs from the top of the wave to the bottom: twice the peak.
2. V_peak = 6 V over 2 = 3 V
Answer: 3 V
Problem 2 of 10
One cycle of a signal takes 2.5 ms. What is its frequency?
Show the working
1. Use f = 1 over T, with T in seconds: 2.5 ms = 0.0025 s.
2. f = 1 over 0.0025 s = 400 Hz
Answer: 400 Hz
Problem 3 of 10
A tuning note is 440 Hz. How long does one cycle take, in milliseconds?
Show the working
1. Use T = 1 over f.
2. T = 1 over 440 Hz ≈ 0.00227 s = 2.27 ms
Answer: 2.27 ms
Problem 4 of 10
A sine wave peaks at 12 V. What is its RMS voltage?
Show the working
1. For a sine, V_rms = V_peak over √2, with √2 ≈ 1.414.
2. V_rms = 12 V over 1.414 ≈ 8.49 V RMS
Answer: 8.49 V
Problem 5 of 10
A transformer's output is rated 24 V RMS. What is its peak voltage?
Show the working
1. Going back from RMS to the peak, multiply by √2 ≈ 1.414.
2. V_peak = 24 V × 1.414 ≈ 33.9 V
Answer: 33.9 V
Problem 6 of 10
A 9 V RMS sine wave drives an 18 Ω heater. How much power does the heater take?
Show the working
1. RMS values work in the power formulas just as DC does: P = V_rms² over R.
2. P = 9² over 18 = 81 over 18 = 4.5 W
Answer: 4.5 W
Problem 7 of 10
A 10 V peak sine wave sits across a 50 Ω resistor. How much power does the resistor turn into heat?
Show the working
1. Turn the peak into RMS first: V_rms = 10 V over √2 ≈ 7.07 V.
2. P = V_rms² over R = 7.07² over 50 ≈ 50 over 50 = 1 W. Using the peak would give 2 W, twice too much.
Answer: 1 W
Problem 8 of 10
A transformer has 800 turns on its primary and 60 on its secondary. It runs from 120 V RMS mains. What is its output voltage?
Show the working
1. The voltage scales by the turns ratio: V_out over V_in = N_out over N_in.
2. V_out = 120 V × 60 over 800 = 9 V RMS
Answer: 9 V
Problem 9 of 10
A transformer steps 230 V RMS mains down 20 to 1. What peak voltage does its output reach?
Show the working
1. The turns ratio scales the RMS voltage: 230 V over 20 = 11.5 V RMS.
2. Multiply by √2 for the peak: 11.5 V × 1.414 ≈ 16.3 V peak
Answer: 16.3 V
Problem 10 of 10
An ideal transformer turns 230 V RMS into 12 V RMS and delivers 2 A RMS to a load. What current does it draw from the mains, in milliamps?
Show the working
1. An ideal transformer wastes nothing, so the power in equals the power out.
2. Power out: P = 12 V × 2 A = 24 W.
3. Current in: I = 24 W over 230 V ≈ 0.104 A = 104 mA, about a nineteenth of the 2 A on the low-voltage side.
Answer: 104 mA

Build it in a free lesson

“Sine waves” is lesson 1 of 6 in Oscillation Coast, Electronics World 4. Free with an account. Lessons unlock in order, so you’ll do the 21 before it first.

Free lesson

Now build it yourself

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

“Sine waves” is lesson 1 of 6 in Oscillation Coast, Electronics World 4. Free with an account. Lessons unlock in order, so you’ll do the 21 before it first.

Or start from the very first lesson, no account needed

The ideas behind it

Questions people ask

What's the difference between peak, peak-to-peak and RMS?

Peak is the highest voltage, measured from zero. Peak-to-peak is top to bottom, twice the peak. RMS is the steady DC voltage that would heat a resistor just as much: for a sine, the peak ÷ √2. Mains (230 V or 120 V, depending on your country) and multimeters quote RMS.

Which voltage goes into P = V² ÷ R for AC?

The RMS. With RMS values the power formulas work exactly as they do for DC. Put the peak in and you get twice the true power.

Does a transformer make extra power?

No. An ideal one passes the power straight through. Step the voltage down and the current steps up by the same ratio, so V × I is the same on both sides. A real transformer loses a little as heat in its coils and core.