Exercises
(7.1) 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?
Answer
Given: R = 100 Ω, V = 220 V (rms), ν = 50 Hz.
(a) I = V/R → I = 220/100 → I = 2.2 A
(b) For a pure resistor the power factor is 1, so P = VI
⇒ P = 220 × 2.2
⇒ P = 484 W
(7.2) (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?
Answer
(a) V = v_m/√2
⇒ V = 300/√2
⇒ V ≈ 212.1 V
(b) i_m = √2 I
⇒ i_m = 1.414 × 10
⇒ i_m ≈ 14.1 A
(7.3) A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer
Given: L = 44 mH = 44 × 10⁻³ H, V = 220 V, ν = 50 Hz.
Inductive reactance X_L = 2πνL
⇒ X_L = 2 × 3.14 × 50 × 44 × 10⁻³
⇒ X_L ≈ 13.8 Ω
I = V/X_L → I = 220/13.8 → I ≈ 15.9 A
(7.4) A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Answer
Given: C = 60 μF = 60 × 10⁻⁶ F, V = 110 V, ν = 60 Hz.
Capacitive reactance X_C = 1/(2πνC)
⇒ X_C = 1/(2 × 3.14 × 60 × 60 × 10⁻⁶)
⇒ X_C ≈ 44.2 Ω
I = V/X_C → I = 110/44.2 → I ≈ 2.49 A
(7.5) In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Answer
Average power P = VI cos φ.
For a pure inductor the current lags the voltage by π/2, so φ = π/2 and cos φ = 0
⇒ P = 0.
For a pure capacitor the current leads the voltage by π/2, so again cos φ = 0
⇒ P = 0.
In both cases the net power absorbed over a full cycle is zero: energy taken from the source in one quarter cycle (stored in the magnetic field of the inductor or the electric field of the capacitor) is returned to the source in the next quarter cycle.
(7.6) A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer
Given: C = 30 × 10⁻⁶ F, L = 27 × 10⁻³ H.
ω₀ = 1/√(LC)
⇒ ω₀ = 1/√(27 × 10⁻³ × 30 × 10⁻⁶)
⇒ ω₀ = 1/√(8.1 × 10⁻⁷) = 1/(9 × 10⁻⁴)
⇒ ω₀ ≈ 1.1 × 10³ rad s⁻¹
(7.7) A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Answer
When the supply frequency equals the natural frequency, the circuit is in resonance: X_L = X_C, so Z = R = 20 Ω and cos φ = 1.
I = V/Z → I = 200/20 → I = 10 A
P = I²R → P = (10)² × 20 → P = 2000 W
(7.8) Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Answer
Given: V = 230 V (rms), L = 5.0 H, C = 80 × 10⁻⁶ F, R = 40 Ω.
(a) ω₀ = 1/√(LC)
⇒ ω₀ = 1/√(5.0 × 80 × 10⁻⁶) = 1/√(4 × 10⁻⁴)
⇒ ω₀ = 50 rad s⁻¹
ν₀ = ω₀/2π
⇒ ν₀ = 50/(2 × 3.14)
⇒ ν₀ ≈ 7.96 Hz
(b) At resonance X_L = X_C, so Z = R = 40 Ω.
rms current I = V/Z
⇒ I = 230/40 = 5.75 A
⇒ amplitude i_m = √2 I = 1.414 × 5.75 ≈ 8.13 A
(c) X_L = ω₀L = 50 × 5.0 = 250 Ω; X_C = 1/(ω₀C) = 1/(50 × 80 × 10⁻⁶) = 250 Ω.
V_R = IR
⇒ V_R = 5.75 × 40
⇒ V_R = 230 V
V_L = IX_L
⇒ V_L = 5.75 × 250
⇒ V_L ≈ 1437.5 V
V_C = IX_C
⇒ V_C = 5.75 × 250
⇒ V_C ≈ 1437.5 V
The voltages across L and C are equal in magnitude but 180° out of phase (V_L leads I by 90°, V_C lags I by 90°). Hence V_LC = I(X_L − X_C) = 5.75 × (250 − 250) = 0. The whole source voltage (230 V) appears across R.
Important Extra Questions
Why is the rms value of an alternating current used rather than its average value?
Answer
The average value of a sinusoidal current over a complete cycle is zero, so it says nothing about its effect. The rms value I = i_m/√2 ≈ 0.707 i_m equals the steady dc current that would produce the same average heating in a resistor, so it is the meaningful measure (ac meters read rms values).
What is meant by power factor? When is it zero and when is it one?
Answer
In P = VI cos φ, the factor cos φ = R/Z is called the power factor. It is zero for a purely inductive or purely capacitive circuit (φ = ±π/2) and equals one for a purely resistive circuit or a series LCR circuit at resonance (φ = 0).
What is wattless current?
Answer
The component of current that is 90° out of phase with the voltage (I sin φ) contributes no average power, since its power factor is zero. It is called wattless current.
Why is resonance not possible in an RL or RC circuit?
Answer
Resonance requires X_L and X_C to cancel each other, which needs both an inductor and a capacitor. In an RL or RC circuit only one reactance is present, so the impedance never reduces to R and no resonance occurs.
A transformer has 100 turns in the primary and 2000 turns in the secondary. If the primary is connected to 220 V ac, find the secondary voltage. Is it step-up or step-down?
Answer
V_s/V_p = N_s/N_p
⇒ V_s = 220 × (2000/100)
⇒ V_s = 4400 V. Since N_s > N_p, it is a step-up transformer (the current in the secondary is reduced in the same ratio for an ideal transformer).