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Question

What is the effect on an ideal transformer’s secondary voltage if the number of turns in the secondary
coil is halved?

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Explanation

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. In an ideal transformer, (V_s/V_p) = (N_s/N_p) . If the number of secondary turns ( N_s ) is halved, the secondary voltage ( V_s ) becomes half its original value, assuming primary voltage ( V_p ) remains constant. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives It halves, consistent with phasor

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