A voltage supply of 12V r.m.s. and frequency 90Hz is connected to a 4 ω resistor. Calculate the peak value of the current

Analyze the question: An AC supply with an r.m.s. voltage of 12 V is connected to a 4 Ω resistor. We need the peak current. The frequency (90 Hz) is not needed for a purely resistive circuit, so it is a distractor.

Concept:

• For a sinusoidal AC supply: V_peak = √2 × V_rms
• Ohm's law applies to a resistor: I = V/R
• So I_peak = V_peak / R = √2 × V_rms / R

Solution:

1. Find the r.m.s. current: I_rms = V_rms / R = 12 / 4 = 3.0 A

2. Convert to peak current: I_peak = √2 × I_rms = 1.414 × 3.0 ≈ 4.24 A

So the peak current is about 4.2 A.

Checking the options:

• E (3.0 A) is the r.m.s. current, which is a trap if you forget to convert to peak.
• D (4.2 A) matches our calculation.

Exam tip: Whenever you see "r.m.s." and "peak" in the same question, remember that peak = √2 × r.m.s. (about 1.414). Also, in a purely resistive circuit the frequency doesn't affect the current. Examiners often include it just to distract you.