Questions & explanations
1. A transformer is connected to an AC source. When the secondary is open, the primary current is 0.5 A. When a load drawing 2 A on the secondary is connected, what is the approximate primary current? (Assume ideal transformer with turns ratio 10:1)
- 0.5 A
- 20 A
- 2.5 A
- 0.7 A
Answer: 0.7 A
In an ideal transformer, the primary current has two components: magnetizing current (0.5 A) and load component. The load component is I_s / turns ratio = 2 A / 10 = 0.2 A. Total primary current = √(0.5² + 0.2²) ≈ 0.54 A, but since the magnetizing current is small and the load component adds in quadrature, the approximate total is about 0.7 A. The core flux remains constant due to Lenz's law.
2. A 10 kW inductive load at 230 V, 50 Hz has power factor 0.6 lagging. What capacitance in parallel raises the power factor to 0.9 lagging?
- 520 μF
- 260 μF
- 1040 μF
- 130 μF
Answer: 520 μF
The required capacitance is C = P(tan φ1 − tan φ2) / (ω V_rms²). For cos φ1 = 0.6, tan φ1 = 4/3. For cos φ2 = 0.9, tan φ2 = √(1−0.81)/0.9 = 0.484. So tan φ1 − tan φ2 = 0.8493. P = 10×10³ W, ω = 2π×50 = 314 rad/s, V_rms = 230 V. Then C = 10⁴ × 0.8493 / (314 × 230²) = 8493 / (314 × 52900) ≈ 8493 / 16610600 ≈ 5.11×10⁻⁴ F = 511 μF ≈ 520 μF.
3. In an LR circuit with R=4 Ω, L=2 H, and V₀=12 V, find the heat dissipated in the resistor up to time τ (time constant).
- 18.0 J
- 3.03 J
- 36.0 J
- 9.0 J
Answer: 3.03 J
τ = L/R = 0.5 s. Current i(t) = (V₀/R)(1 - e^{-t/τ}) = 3(1 - e^{-2t}) A. Heat Q = ∫₀^τ i²R dt = 4∫₀^0.5 9(1 - e^{-2t})² dt = 36∫₀^0.5 (1 - 2e^{-2t} + e^{-4t}) dt = 36[t + e^{-2t} - (1/4)e^{-4t}]₀^0.5 = 36[(0.5 + e^{-1} - 0.25e^{-2}) - (0 + 1 - 0.25)] = 36(0.5 + 0.3679 - 0.0338 - 0.75) = 36(0.0841) ≈ 3.03 J.
4. Two coaxial solenoids have n₁=500 turns/m, n₂=1000 turns/m, area A=0.01 m², length l=0.2 m. They carry currents I₁=2 A and I₂=1 A in aiding sense. Find the total stored energy.
- 5.03 mJ
- 2.51 mJ
- 0 mJ
- 7.54 mJ
Answer: 5.03 mJ
Using principle of energy in coupled inductors: U = ½L₁I₁² + ½L₂I₂² + MI₁I₂ for aiding currents. Compute L₁ = μ₀n₁²Al = 6.283×10⁻⁴ H, L₂ = μ₀n₂²Al = 2.513×10⁻³ H, M = μ₀n₁n₂Al = 1.257×10⁻³ H. Then U = ½×6.283e-4×4 + ½×2.513e-3×1 + 1.257e-3×2×1 = 1.257 mJ + 1.257 mJ + 2.513 mJ = 5.027 mJ ≈ 5.03 mJ.
5. A step-down transformer has N_p = 1000, N_s = 100, V_p = 220 V. The secondary feeds a 10 Ω load. Primary resistance is 2 Ω and iron loss is 200 W. What is the efficiency?
- 19.5%
- 80.5%
- 96.2%
- 24.2%
Answer: 19.5%
V_s = (100/1000)×220 = 22 V. I_s = 22/10 = 2.2 A. P_out = 22×2.2 = 48.4 W. I_p = (100/1000)×2.2 = 0.22 A. Copper loss = (0.22)²×2 = 0.0968 W. Total loss = 0.0968 + 200 = 200.0968 W. P_in = 48.4 + 200.0968 = 248.4968 W. Efficiency = 48.4/248.4968 ≈ 0.1948 = 19.48%, rounded to 19.5%.
6. The I_rms vs ω graph of a series LCR circuit shows peak current 2 A at ω0 = 2000 rad/s. The half-power points are at ω1 = 1900 rad/s and ω2 = 2100 rad/s. If C = 1 μF, what is the value of L?
- 0.10 H
- 0.50 H
- 0.25 H
- 0.05 H
Answer: 0.25 H
At resonance, ω0 = 1/√(LC). Given ω0 = 2000 rad/s and C = 1 μF = 10⁻⁶ F, L = 1/(ω0²C) = 1/(4×10⁶ × 10⁻⁶) = 0.25 H. The half-power bandwidth Δω = 200 rad/s gives Q = ω0/Δω = 10, consistent with L = 0.25 H and R = ω0L/Q = 50 Ω, but L is directly determined by ω0 and C.
7. A step-down transformer has 1000 turns on primary and 100 turns on secondary. Primary is connected to 220 V AC. A 11 Ω bulb is connected across secondary. Find the primary current.
- 0.2 A
- 2 A
- 0.02 A
- 20 A
Answer: 0.2 A
Using transformer turns ratio, secondary voltage V_s = (N_s/N_p) × V_p = (100/1000)×220 = 22 V. Secondary current I_s = V_s/R = 22/11 = 2 A. For ideal transformer, input power equals output power: V_p I_p = V_s I_s, so I_p = (V_s I_s)/V_p = (22×2)/220 = 0.2 A.
8. A series LCR circuit has V_rms = 100 V, L = 0.1 H, C = 10 μF, R = 10 Ω. At a frequency ω = 1256.6 rad/s, what is the average power dissipated?
- 100 W
- 45.0 W
- 212 W
- 10.0 W
Answer: 45.0 W
Average power in an AC circuit is P = V_rms I_rms cos φ. Compute XL = ωL = 125.66 Ω, XC = 1/(ωC) = 79.58 Ω, net reactance X = 46.08 Ω, impedance Z = √(R²+X²) = 47.17 Ω. Then I_rms = V_rms/Z = 2.12 A, cos φ = R/Z = 0.212. So P = 100 × 2.12 × 0.212 = 45.0 W.
9. In an ideal transformer, the phase difference between the primary voltage and the secondary voltage is:
- 0°
- 90°
- 270°
- 180°
Answer: 180°
Both primary and secondary EMFs are given by ε = -N dΦ/dt. For the primary, the applied voltage V_p balances the induced EMF: V_p = -ε_p = N_p dΦ/dt. For the secondary, V_s = ε_s = -N_s dΦ/dt. Thus V_s = -(N_s/N_p) V_p, showing a 180° phase difference.
10. In the transformer symbol, dots are placed on the same side of both windings. If the primary voltage is V_p = V_0 sin(ωt), the secondary voltage is:
- V_s = -(N_s/N_p) V_0 sin(ωt)
- V_s = (N_s/N_p) V_0 sin(ωt)
- V_s = (N_p/N_s) V_0 sin(ωt)
- V_s = -(N_p/N_s) V_0 sin(ωt)
Answer: V_s = -(N_s/N_p) V_0 sin(ωt)
By Lenz's law, the induced secondary voltage opposes the flux change. Dots on same side indicate that when primary current enters the dotted terminal, secondary current leaves the dotted terminal, giving a 180° phase shift. Thus V_s = -(N_s/N_p) V_p.
11. A series LCR circuit has L = 0.1 H, C = 10 μF, R = 50 Ω, and source frequency 50 Hz. What is the phase difference between current and voltage?
- Current leads by about 80°
- Current lags by about 80°
- Current leads by about 10°
- Current lags by about 10°
Answer: Current leads by about 80°
Using phasor analysis, tan φ = (XL - XC)/R. ω = 2πf = 100π ≈ 314 rad/s. XL = ωL = 31.4 Ω, XC = 1/(ωC) ≈ 318.3 Ω. Net reactance X = XL - XC = -286.9 Ω (capacitive). tan φ = -286.9/50 = -5.74, so φ ≈ -80°, meaning current leads voltage by about 80°.
12. In a series LCR circuit, the bandwidth Δω is equal to which of the following?
- R/L
- L/R
- R/(2L)
- 2R/L
Answer: R/(2L)
Bandwidth Δω is the difference between half-power frequencies. At half-power, current is 1/√2 of maximum, so impedance magnitude is √2 R. Solving (ωL - 1/(ωC))² = R² gives ω₂ - ω₁ = R/L. However, bandwidth is defined as Δω = (ω₂ - ω₁)/2 = R/(2L).