Alternating Current

1 exercises10 questions solved

Exercise 7.1AC Circuits — R, L, C and Series LCR

A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?

Solution

R = 100 Ω, V_rms = 220 V, f = 50 Hz (a) rms current: I_rms = V_rms/R = 220/100 = 2.2 A (b) Power consumed: P = V_rms × I_rms × cosφ For a pure resistor: φ = 0°, cosφ = 1 P = 220 × 2.2 × 1 = 484 W

(a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current?

Solution

(a) V_peak = 300 V V_rms = V_peak/√2 = 300/√2 = 300/1.414 ≈ 212.1 V (b) I_rms = 10 A I_peak = I_rms × √2 = 10 × 1.414 ≈ 14.14 A

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Solution

L = 44 mH = 44 × 10⁻³ H, V_rms = 220 V, f = 50 Hz Inductive reactance: X_L = 2πfL = 2π × 50 × 44 × 10⁻³ = 2π × 2.2 = 4.4π = 13.82 Ω rms current: I_rms = V_rms/X_L = 220/13.82 ≈ 15.92 A I_rms ≈ 15.92 A

A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Solution

C = 60 μF = 60 × 10⁻⁶ F, V_rms = 110 V, f = 60 Hz Capacitive reactance: X_C = 1/(2πfC) = 1/(2π × 60 × 60 × 10⁻⁶) = 1/(2π × 3600 × 10⁻⁶) = 1/(22.62 × 10⁻³) = 44.2 Ω rms current: I_rms = V_rms/X_C = 110/44.2 ≈ 2.49 A

In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Solution

For a pure inductor (Exercise 7.3): Phase difference φ = 90° (current lags voltage by 90°) Power = V_rms × I_rms × cosφ = V_rms × I_rms × cos90° = 0 For a pure capacitor (Exercise 7.4): Phase difference φ = 90° (current leads voltage by 90°) Power = V_rms × I_rms × cos90° = 0 Explanation: In a pure inductor or pure capacitor, energy is stored in the magnetic/electric field during one half-cycle and returned to the circuit in the other half-cycle. No net energy is dissipated — the power factor (cosφ) is zero. These are purely reactive elements.

Obtain the resonant frequency ωᵣ of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?

Solution

L = 2.0 H, C = 32 μF = 32 × 10⁻⁶ F, R = 10 Ω Resonant frequency: ωᵣ = 1/√(LC) = 1/√(2.0 × 32 × 10⁻⁶) = 1/√(64 × 10⁻⁶) = 1/(8 × 10⁻³) = 125 rad/s Q-value: Q = ωᵣL/R = 125 × 2.0 / 10 = 250/10 = 25 Resonant frequency = 125 rad/s, Q-value = 25

A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Solution

C = 30 μF = 30 × 10⁻⁶ F, L = 27 mH = 27 × 10⁻³ H Angular frequency: ω = 1/√(LC) = 1/√(27 × 10⁻³ × 30 × 10⁻⁶) = 1/√(810 × 10⁻⁹) = 1/√(8.1 × 10⁻⁷) = 1/(9 × 10⁻⁴ × √(1)) √(8.1 × 10⁻⁷) = √8.1 × 10⁻³·⁵ = 2.846 × 10⁻³·⁵ = 2.846 × 10⁻³ × 10⁻⁰·⁵ ... Let me compute directly: 27 × 10⁻³ × 30 × 10⁻⁶ = 810 × 10⁻⁹ = 8.1 × 10⁻⁷ √(8.1 × 10⁻⁷) = √8.1 × 10⁻³·⁵ ≈ 2.846 × 3.162 × 10⁻⁴ ... Actually: √(8.1 × 10⁻⁷) = √(81 × 10⁻⁸) = 9 × 10⁻⁴ ω = 1/(9 × 10⁻⁴) ≈ 1.11 × 10³ rad/s

Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the natural frequency of the circuit?

Solution

C = 30 × 10⁻⁶ F, Q₀ = 6 mC = 6 × 10⁻³ C, L = 27 × 10⁻³ H Initial energy (all in capacitor initially): U = Q₀²/(2C) = (6 × 10⁻³)² / (2 × 30 × 10⁻⁶) = 36 × 10⁻⁶ / (60 × 10⁻⁶) = 36/60 = 0.6 J Natural frequency: ω = 1.11 × 10³ rad/s (from previous question) f = ω/(2π) = 1.11 × 10³ / (2π) = 1110/6.283 ≈ 176.7 Hz Total energy = 0.6 J, Natural frequency ≈ 177 Hz

Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω. Determine (a) the source frequency which drives the circuit in resonance, (b) the impedance of the circuit and the amplitude of the current at the resonating frequency, (c) the rms potential drop across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Solution

L = 5.0 H, C = 80 μF = 80 × 10⁻⁶ F, R = 40 Ω, V_rms = 230 V (a) Resonant frequency: ω₀ = 1/√(LC) = 1/√(5.0 × 80 × 10⁻⁶) = 1/√(400 × 10⁻⁶) = 1/(20 × 10⁻³) = 50 rad/s (b) At resonance: Z = R = 40 Ω (since X_L = X_C) I_rms = V_rms/Z = 230/40 = 5.75 A I₀ = I_rms × √2 = 5.75 × 1.414 ≈ 8.13 A (c) X_L = ω₀L = 50 × 5.0 = 250 Ω X_C = 1/(ω₀C) = 1/(50 × 80 × 10⁻⁶) = 1/0.004 = 250 Ω V_R = I_rms × R = 5.75 × 40 = 230 V V_L = I_rms × X_L = 5.75 × 250 = 1437.5 V V_C = I_rms × X_C = 5.75 × 250 = 1437.5 V Potential drop across LC combination: V_LC = V_L – V_C = 1437.5 – 1437.5 = 0 V At resonance X_L = X_C, the voltages across L and C are equal but 180° out of phase, so they cancel. V_LC = 0 ✓

Q10

An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. (a) What is the total energy stored initially? Is it conserved during LC oscillations? (b) What is the natural frequency of the circuit? (c) At what time is the energy stored (i) completely electrical (ii) completely magnetic? (d) At what times is the total energy shared equally between the inductor and the capacitor?

Solution

L = 20 × 10⁻³ H, C = 50 × 10⁻⁶ F, Q₀ = 10 × 10⁻³ C (a) Initial energy: U = Q₀²/(2C) = (10⁻²)²/(2 × 50 × 10⁻⁶) = 10⁻⁴/(10⁻⁴) = 1 J Yes, total energy is conserved (no resistance), it oscillates between electric and magnetic forms. (b) Natural frequency: ω = 1/√(LC) = 1/√(20 × 10⁻³ × 50 × 10⁻⁶) = 1/√(10⁻⁶) = 1/10⁻³ = 10³ rad/s f = ω/(2π) = 1000/(2π) ≈ 159 Hz Time period T = 1/f = 2π/ω = 2π × 10⁻³ ≈ 6.28 × 10⁻³ s (c)(i) Energy completely electrical (all in capacitor, Q = maximum): This happens at t = 0, T/2, T, 3T/2, ... = nT/2 t = 0, π × 10⁻³ s, 2π × 10⁻³ s, ... (c)(ii) Energy completely magnetic (all in inductor, Q = 0, I = maximum): This happens at t = T/4, 3T/4, ... = (2n+1)T/4 t = T/4 = π/2 × 10⁻³ ≈ 1.57 × 10⁻³ s, then every T/2 after that (d) Energy equally shared: Electrical energy = ½U → Q²/(2C) = U/2 → Q = Q₀/√2 This happens at ωt = π/4, 3π/4, ... t = T/8, 3T/8, 5T/8, 7T/8, ... t = (2n+1)T/8 = (2n+1)π/(4ω) = (2n+1) × (π/4) × 10⁻³ s

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