Electromagnetic Induction and Alternating Current — Practice Quiz

ε₀ = NBAω

Step 1: The magnetic flux through a rotating coil: φ(t) = NBA cos(ωt). Step 2: By Faraday's Law for N turns: ε = -N dφ/dt. Step 3: Differentiating: ε = -N × (-NBAω sin(ωt)) — wait, for a single-turn flux NBA cosωt, ε = NBAω sin(ωt). Step 4: Therefore, the peak (maximum) EMF is ε₀ = NBAω. Step 5: Option A (NBA) is just the maximum flux, not the EMF. Option C (NBA/ω) confuses multiplication and division. Option D adds an extra N, which is incorrect.

0.6 A

Step 1: For an ideal transformer: Np/Ns = Ip/Is (turns ratio equals inverse current ratio). Step 2: Given Np = 100, Ns = 500, Ip = 3 A. Step 3: Is = Ip × (Np/Ns) = 3 × (100/500) = 3 × 0.2 = 0.6 A. Step 4: Verification using power: Primary power = 120 × 3 = 360 W. Secondary voltage = 120 × (500/100) = 600 V. Secondary power = 600 × 0.6 = 360 W ✓ (ideal transformer, no power loss). Step 5: Option A (15 A) multiplies instead of dividing. Option B (3 A) ignores the turns ratio. Option D (1.5 A) uses an incorrect ratio calculation.

Current leads voltage by 90°

Step 1: For a capacitor, charge q = CV = CV_m cos(ωt). Step 2: Current I = dq/dt = -CV_m ω sin(ωt). Step 3: Since -sin(ωt) = cos(ωt + 90°), the current can be written as I = CV_m ω cos(ωt + 90°). Step 4: This shows the current is ahead of the voltage by 90°, i.e., current LEADS voltage by 90°. Step 5: Option A describes an inductor (current lags). Option B describes a pure resistor. Option D (45°) is incorrect — the phase difference for a pure capacitor is always exactly 90°.

311 V

Step 1: The relationship between peak and rms values: V_rms = V_m / √2. Step 2: Rearranging: V_m = V_rms × √2. Step 3: Substituting: V_m = 220 × √2 = 220 × 1.414 ≈ 311 V. Step 4: This is why AC mains at 220 V (rms) can be dangerous — the peak voltage reaches about 311 V. Step 5: Option A (220 V) confuses rms with peak. Option B (155.6 V) divides by √2 instead of multiplying. Option D (440 V) doubles the rms value, which is incorrect.