Chapter 7: Alternating Current
Key Formulas:
• V = V₀ sin ωt, I = I₀ sin(ωt − φ)
• Impedance: Z = √(R² + (XL − XC)²)
• XL = ωL, XC = 1/(ωC)
• Resonance: XL = XC, f = 1/(2π√LC)
• Power: P = VI cosφ
Q1. An AC source has ω = 300 rad/s. Find the value of L and C for series resonance at 500 Hz.
Solution
At resonance: ω₀ = 1/√(LC)
300 = 1/√(LC)
If we assume f = 500 Hz, then ω = 2π(500) = 3141.6 rad/s
For ω = 300: LC = 1/300² = 1/90000
For a given L, C = 1/(L × 90000)
If L = 0.1 H: C = 1/(9000) ≈ 111 μF
Q2. In an LCR circuit with R = 20Ω, L = 0.5H, C = 100μF, connected to 220V, 50Hz supply, find (i) impedance (ii) current (iii) phase angle.
Solution
ω = 2π(50) = 314 rad/s
XL = ωL = 314 × 0.5 = 157 Ω
XC = 1/(ωC) = 1/(314 × 10−4) = 31.85 Ω
(i) Z = √(20² + (157 − 31.85)²) = √(400 + 15663) = √16063 = 126.7 Ω
(ii) I = V/Z = 220/126.7 = 1.74 A
(iii) tanφ = (XL − XC)/R = 125.15/20 = 6.26
φ = 80.9° (inductive)
Q3. A transformer steps down 220V to 22V. If the primary coil has 1000 turns, find the number of turns in the secondary.
Solution
V₁/V₂ = N₁/N₂
220/22 = 1000/N₂
N₂ = 1000/10 = 100 turns
Q4. Find the power dissipated in a pure inductor of inductance 0.5 H connected to an AC source of 200V, 50Hz.
Solution
In a pure inductor, phase angle φ = 90°
Power P = VI cos 90° = 0 W
The average power dissipated in a pure inductor is zero because it only stores and releases energy.
Q5. Why is the power factor important in AC circuits? What is its ideal value?
Solution
Power factor = cosφ = R/Z
Power factor determines the ratio of actual power (dissipated) to apparent power.
Ideal value = 1 (purely resistive circuit)
When cosφ = 1, maximum power is delivered to the load. Low power factor means more current is needed for the same power, causing higher losses in transmission lines.
Q6. An AC voltage v = 200 sin 100πt is applied across a series LCR circuit with R = 20 Ω, L = 0.1 H, C = 100 μF. Find (a) impedance (b) peak current (c) phase angle.
Solution
ω = 100π = 314 rad/s
XL = ωL = 314 × 0.1 = 31.4 Ω
XC = 1/(ωC) = 1/(314 × 10−4) = 31.85 Ω
(a) Z = √(R² + (XL − XC)²) = √(400 + 0.2025) ≈ 20 Ω
(b) I₀ = V₀/Z = 200/20 = 10 A
(c) tanφ = (XL − XC)/R = −0.45/20 ≈ −0.0225
φ ≈ −1.29° (slightly capacitive)
Q7. What is resonance in a series LCR circuit? Derive the expression for resonant frequency.
Solution
At resonance, XL = XC, so the impedance is minimum (Z = R) and current is maximum.
ωL = 1/(ωC)
ω² = 1/(LC)
ω₀ = 1/√(LC)
f₀ = 1/(2π√LC)
Q8. A transformer has 1000 turns in primary and 200 turns in secondary. If the primary voltage is 220V and primary current is 1A, find (i) secondary voltage (ii) secondary current (iii) power output.
Solution
(i) V₁/V₂ = N₁/N₂
220/V₂ = 1000/200 = 5
V₂ = 44 V
(ii) Assuming ideal transformer: V₁I₁ = V₂I₂
220 × 1 = 44 × I₂
I₂ = 5 A
(iii) P = V₁I₁ = 220 × 1 = 220 W
Q9. An inductor of inductance 2H, a capacitor of capacitance 20 μF, and a resistor of resistance 10 Ω are connected in series to an AC source of emf e = 100 sin 100t V. Find (i) the impedance at resonance (ii) the current at resonance (iii) the power dissipated.
Solution
ω = 100 rad/s
Resonant frequency: ω₀ = 1/√(LC) = 1/√(2 × 20 × 10−6) = 1/√(4 × 10−5) = 158.1 rad/s
At ω = 100 (not at resonance):
XL = 100 × 2 = 200 Ω
XC = 1/(100 × 20 × 10−6) = 500 Ω
Z = √(10² + (200−500)²) = √(100 + 90000) ≈ 300.2 Ω
(i) At resonance Zmin = 10 Ω
(ii) I₀ = V₀/R = 100/10 = 10 A
(iii) P = I₀²R/2 = (100)(10)/2 = 500 W
Q10. Distinguish between DC and AC. Why is AC preferred for long-distance power transmission?
Solution
DC: Current flows in one direction only. Constant magnitude.
AC: Current reverses direction periodically. Varies sinusoidally.
AC is preferred because:
• Can be stepped up/down using transformers
• At high voltage, transmission losses (I²R) are reduced
• AC generators are simpler than DC generators
• No commutator needed
Q11. Draw the phasor diagram for (i) a pure R-L circuit (ii) a pure R-C circuit. In each case, show the phase relationship between voltage and current.
Solution
(i) R-L circuit: V leads I by angle φ where tanφ = XL/R.
Phasor diagram: I along x-axis, VR along x-axis, VL along y-axis, V is the resultant.
(ii) R-C circuit: I leads V by angle φ where tanφ = XC/R.
Phasor diagram: V along x-axis, VR along x-axis, VC along −y-axis, I is the resultant.
Q12. The RMS value of an AC voltage is 220V. Find (i) the peak value (ii) the average value over a complete cycle.
Solution
(i) V₀ = Vrms × √2 = 220 × 1.414 = 311 V
(ii) Average value over a complete cycle = 0 V
Note: The average over a half-cycle is 2V₀/π = 0.637 V₀ = 198 V
Q13. A 100 Ω resistor, a 1 H inductor, and a 10 μF capacitor are connected in series to a 100V, 50 Hz AC supply. Find (i) the impedance (ii) the current (iii) the voltage across the capacitor.
Solution
ω = 2π(50) = 314 rad/s
XL = 314 × 1 = 314 Ω
XC = 1/(314 × 10−5) = 318.3 Ω
(i) Z = √(100² + (314 − 318.3)²) = √(10000 + 18.49) ≈ 100.1 Ω
(ii) I = V/Z = 100/100.1 ≈ 0.999 A ≈ 1 A
(iii) VC = IXC = 1 × 318.3 = 318.3 V (can exceed source!)