A-Level Magnetic flux, the induction laws of Faraday and Lenz, and power transformers: worked solution

Question

For that ideal transformer (2700 : 450 turns), the primary e.m.f. reads E = 220 sin(100πt) volts, t in seconds. Obtain the secondary coil's induced r.m.s. e.m.f. [3]

Principle: the transformer equation V_s / V_p = N_s / N_p holds for an ideal transformer, and it compares like with like - both voltages must be r.m.s. values (or both peak values). The e.m.f. we are given is written as an instantaneous equation, so the first job is to extract an r.m.s. value from it. Step 1 - read the peak value out of the given equation. A sinusoidal e.m.f. is written E = E₀ sin(ωt), where E₀ is the peak (maximum) value. Comparing with E = 220 sin(100πt), the number in front of the sine is the peak: E₀ = 220 V (The 100π is the angular frequency, ω = 100π rad s⁻¹, which we do not need here.) Step 2 - convert the peak value to an r.m.s. value. For a sinusoid, V_rms = V₀ / √2. Primary V_rms = 220 / √2 Primary V_rms = 220 / 1.414 Primary V_rms = 155.6 V Step 3 - apply the transformer equation. V_s / V_p = N_s / N_p, so multiply both sides by V_p: V_s = (N_s / N_p) × V_p The turns are 2700 on the primary and 450 on the secondary, so N_s / N_p = 450 / 2700 = 1/6 (this is a step-down transformer, so we expect a smaller output) Step 4 - substitute. V_s(r.m.s.) = (1/6) × 155.6 V V_s(r.m.s.) = 25.9 V The secondary r.m.s. e.m.f. is about 26 V.