Calculate the peak voltage that could be obtained from a 220 V$_{r.m.s}$main supply.

Analyze the question: You are given the r.m.s. (root mean square) voltage of an AC supply and asked for the peak (maximum) voltage.

Concept: For a sinusoidal AC supply, the r.m.s. value is the peak value divided by √2:

V_rms = V_peak / √2

So: V_peak = V_rms × √2

Solution:

V_peak = 220 × √2 V_peak = 220 × 1.4142 V_peak ≈ 311.1 V

Checking the options:

• A (70.7 V) and B (110 V) are too small. The peak must be larger than the r.m.s. value.
• C (155.6 V) is 220 ÷ √2, which is the common mistake of dividing instead of multiplying.
• E (440 V) is 220 × 2, which would be the peak-to-peak value only if the peak were 220 V, and it is not the answer here.
• D (311.1 V) matches our calculation.

Exam tip: The peak is always bigger than the r.m.s. value. Remember "r.m.s. is the smaller one", so multiply by √2 (about 1.414) to go from r.m.s. to peak. For the Nigerian 220 V mains, the peak is about 311 V. Also, the Nigerian mains supply is 50 Hz, a fact that sometimes appears in related questions.