NCERT Solutions: Alternating Current
Page No. 200
Q7.1: 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?
Ans: Resistance of the resistor, R = 100 Ω
Supply voltage, V = 220 V
Frequency, ν = 50 Hz
(a) The rms value of current in the circuit is given as:
Irms = Vrms/R
Substituting the values:
Irms = 220 / 100 = 2.20 A
Thus the rms current is 2.20 A
(b) The net (average) power consumed by a pure resistor in an ac circuit is
P = Vrms Irms cos Φ
For a pure resistor the voltage and current are in phase, so cos Φ = 1.
Therefore:
P = 220 × 2.20 × 1 = 484 W
Hence, the net power consumed over a full cycle is 484 W.
Q7.2: (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?
Ans: (a) Peak voltage of the ac supply, V0 = 300 V
For a sinusoidal quantity:
Vrms = V0/√2
Substituting the value:
Vrms = 300 / √2 ≈ 300 / 1.414 = 212.13 V
Hence, the rms voltage is 212.13 V
(b) The rms value of current is given as:
Irms = 10 A
Peak current is related by:
I0 = Irms √2 = 10 × 1.414 = 14.14 A
Hence, the peak current is 14.14 A
Q7.3: A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Ans: Inductance of inductor, L = 44 mH = 44 × 10-3 H
Supply voltage, V = 220 V
Frequency, ν = 50 Hz
Angular frequency, ω= 2πν = 2π × 50 = 100π rad/s
Inductive reactance, XL = ωL = (100π)(44 × 10-3) Ω
Evaluating:
ω ≈ 314.16 rad/s
XL ≈ 314.16 × 0.044 = 13.82 Ω
Rms value of current is given as:
Irms = Vrms / XL = 220 / 13.82 ≈ 15.92 A
Hence, the rms value of current in the circuit is 15.92 A
Q7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Ans: Capacitance of capacitor, C = 60 μF = 60 × 10-6 F
Supply voltage, V = 110 V
Frequency, ν = 60 Hz
Angular frequency, ω= 2πν = 2π × 60 = 120π rad/s
Capacitive reactance, XC = 1 / (ωC)
Evaluating:
ω ≈ 376.99 rad/s
XC = 1 / (376.99 × 60 × 10-6) ≈ 1 / 0.022619 = 44.24 Ω
Rms value of current is given as:
Irms = Vrms / XC = 110 / 44.24 ≈ 2.49 A
Hence, the rms value of current is 2.49 A
Q7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Ans: In the inductive circuit,
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Net (average) power absorbed by any ac circuit is given by
P = Vrms Irms cos Φ
Where Φ is the phase difference between V and I.
In a pure inductive circuit the current lags the voltage by 90°, so Φ = 90°, and cos 90° = 0.
Therefore, P = 0. The net power absorbed over a full cycle is zero.
In the capacitive circuit,
Rms value of current, I = 2.49 A
Rms value of voltage, V = 110 V
In a pure capacitive circuit the current leads the voltage by 90°, so Φ = 90° and cos 90° = 0.
Therefore, P = 0. The net power absorbed over a full cycle is zero.
Note: In both pure inductive and pure capacitive circuits no net energy is dissipated; energy is exchanged back and forth between the source and the reactive element during each cycle, giving zero average power over a complete cycle.
Q7.6: A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Ans: Capacitance, C = 30 μF = 30 × 10-6 F
Inductance, L = 27 mH = 27 × 10-3 H
Angular frequency of free (natural) oscillations is:
ω = 1 / √(LC)
Evaluating:
LC = (27 × 10-3)(30 × 10-6) = 8.1 × 10-7
√(LC) = 9.0 × 10-4
ω = 1 / (9.0 × 10-4) = 1.11 × 103 rad/s
Hence, the angular frequency of free oscillations is 1.11 × 103 rad/s
Q7.7: 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?
Ans: At resonance the reactive parts cancel and the impedance of the series LCR circuit equals the resistance R.
Resistance, R = 20 Ω
Inductance, L = 1.5 H
Capacitance, C = 35 μF = 35 × 10-6 F
AC supply (rms) voltage, Vrms = 200 V
Impedance at resonance, Z = R = 20 Ω.
Rms current at resonance:
Irms = Vrms / R = 200 / 20 = 10 A
The average power transferred (absorbed) at resonance is:
P = Vrms Irms cos Φ
At resonance voltage and current are in phase, so cos Φ = 1.
P = 200 × 10 = 2000 W
Hence, the average power transferred in one complete cycle at resonance is 2000 W
Q7.8: Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, 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.
Ans: Inductance of the inductor, L = 5.0 H
Capacitance of the capacitor, C = 80 μF = 80 × 10-6 F
Resistance of the resistor, R = 40 Ω
Supply (rms) voltage, V = 230 V
(a) Resonance angular frequency is:
ω0 = 1 / √(LC)
Evaluating:
LC = 5.0 × 80 × 10-6 = 4.0 × 10-4
√(LC) = 2.0 × 10-2
ω0 = 1 / (2.0 × 10-2) = 50 rad/s
Hence, the circuit resonates at ω = 50 rad/s
(b) At resonance the inductive and capacitive reactances cancel each other, so the
impedance equals the resistance:
Z = R = 40 Ω
Rms current at resonance:
Irms = Vrms / R = 230 / 40 = 5.75 A
Amplitude (peak) of the current:
I0 = Irms √2 ≈ 5.75 × 1.414 = 8.13 A
Hence, at resonance the impedance is 40 Ω, the rms current is 5.75 A and the current amplitude is 8.13 A.
(c) Rms potential drop across the resistor:
VR,rms = Irms R = 5.75 × 40 = 230 V
Rms potential drop across the inductor:
VL,rms = Irms × ωL = 5.75 × (50 × 5.0) = 5.75 × 250 = 1437.5 V
Rms potential drop across the capacitor:
VC,rms = Irms × (1 / (ωC)) = 5.75 × 250 = 1437.5 V
At resonance the magnitudes of VL and VC are equal but they are 180° out of phase, so their algebraic sum across the LC combination is zero:
VLC = VL - VC = 1437.5 - 1437.5 = 0 V
Thus, the potential drop across the LC combination is zero at resonance, while the whole supply voltage appears across the resistor (230 V).
Old NCERT Solutions
Q1. 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 (i.e., stored in the capacitor)?
(ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Ans. Inductance of the inductor, L = 20 mH = 20 × 10-3 H
Capacitance of the capacitor, C = 50 μF = 50 × 10-6 F
Initial charge on the capacitor, Q = 10 mC = 10 × 10-3 C
(a) Total energy stored initially in the circuit is given as:
The initial energy resides entirely in the charged capacitor and is
U = Q² / (2C)
= (10 × 10-3)² / [2 × (50 × 10-6)]
= (1 × 10-4) / (1 × 10-4) = 1 J
Hence, the initial total energy is 1 J. Since the resistance is negligible, this energy is conserved and is exchanged between the capacitor and the inductor during LC oscillations.
(b) Natural (angular) frequency of the circuit is
ω = 1 / √(LC)
Evaluating:
LC = (20 × 10-3)(50 × 10-6) = 1.0 × 10-6
√(LC) = 1.0 × 10-3
ω = 1 / (1.0 × 10-3) = 103 rad/s
Hence, ω = 10³ rad/s
(c) (i) The charge on the capacitor during oscillation may be written q(t) = Q cos(ωt).
The capacitor stores maximum electrical energy when q = ±Q, which occurs when cos(ωt) = ±1, i.e. when ωt = n·2π, n = 0, 1, 2, ...
Therefore times when energy is completely electrical:
t = n·(2π / ω) = nT, where T = 2π / ω.
Thus, at t = nT the energy is completely electrical (entirely in the capacitor).
(ii) Magnetic energy is maximum when the capacitor charge is zero (q = 0) because then all energy is in the inductor. q = 0 when cos(ωt) = 0, i.e. ωt = π/2 + nπ.
Hence times when energy is completely magnetic are:
t = (π/2 + nπ) / ω = ( (2n + 1)π / 2 ) / ω.
Thus, at these times the energy is entirely magnetic.
(d) The energies are equal when the capacitor energy = inductor energy = U/2. For the capacitor energy to be half its maximum we need |q| = Q/√2, i.e. cos(ωt) = ±1/√2.
That occurs when ωt = π/4 + nπ/2.
Hence the times are:
t = (π/4 + nπ/2) / ω, where n = 0, 1, 2, ...
At these times the total energy is shared equally between the two elements.
(e) If a resistor is inserted, the oscillations are damped and ultimately all the initial energy is dissipated as heat in the resistor. Thus the amount of energy eventually dissipated equals the initial stored energy, 1 J.
Q2. A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Ans. Inductance of the inductor, L = 0.50 H
Resistance of the resistor, R = 100 Ω
Potential of the supply voltage, V = 240 V (rms)
Frequency of the supply, ν = 50 Hz
(a) Peak (amplitude) supply voltage:
V0 = Vrms √2 = 240 × 1.414 ≈ 339.41 V
Angular frequency:
ω = 2πν = 2π × 50 = 100π ≈ 314.16 rad/s
Inductive reactance:
XL = ωL = 314.16 × 0.50 = 157.08 Ω
Impedance of the coil (series R and L):
Z = √(R² + XL²) = √(100² + 157.08²) ≈ 186.2 Ω
Rms current:
Irms = Vrms / Z = 240 / 186.2 ≈ 1.289 A
Maximum (peak) current:
I0 = Irms √2 ≈ 1.289 × 1.414 = 1.82 A
Hence, the maximum current in the coil is approximately 1.82 A.
(b) Voltage and current in an RL circuit are not in phase. The phase difference Φ (voltage leads current) is given by:
Φ = arctan(XL / R) = arctan(157.08 / 100) ≈ 57.99° = 1.012 rad
The time lag between the voltage maximum and current maximum is Δt = Φ / ω = 1.012 / 314.16 ≈ 0.00322 s = 3.22 ms
Hence, the time lag is about 3.22 ms.
Q3. Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Ans. Inductance of the inductor, L = 0.50 H
Resistance of the resistor, R = 100 Ω
Potential of the supply voltages, V = 240 V (rms)
Frequency of the supply,ν = 10 kHz = 104 Hz
Angular frequency, ω = 2πν= 2π × 104 ≈ 62831.9 rad/s
(a) Peak supply voltage:
V0 = Vrms √2 ≈ 240 × 1.414 = 339.41 V
Inductive reactance:
XL = ωL ≈ 62831.9 × 0.50 = 31415.95 Ω
Impedance:
Z = √(R² + XL²) ≈ √(100² + 31415.95²) ≈ 31416.11 Ω
Rms current:
Irms = Vrms / Z ≈ 240 / 31416.11 ≈ 7.64 × 10-3 A = 0.00764 A
Peak current I0 = Irms √2 ≈ 0.0108 A
Thus the current is extremely small at high frequency.
(b) Phase angle Φ = arctan(XL/R) ≈ arctan(31415.95/100) ≈ 89.82°, nearly 90°. The time lag Δt = Φ / ω ≈ 1.5684 / 62831.9 ≈ 2.50 × 10-5 s (≈ 25 μs).
Conclusion: At very high frequency XL becomes very large, so the impedance of the inductor is very large and the current through the circuit is nearly zero - the inductor behaves like an open circuit for high-frequency signals. In a dc circuit after steady state (frequency effectively zero), XL = ωL = 0 and the inductor behaves like a short circuit (pure conductor), allowing steady current to pass with only the DC resistance limiting it.
Q4. A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Ans. Capacitance of the capacitor, C = 100 μF = 100 × 10-6 F
Resistance of the resistor, R = 40 Ω
Supply voltage (rms), V = 110 V
(a) Frequency of supply,ν = 60 Hz
Angular frequency, ω = 2πν = 2π × 60 ≈ 376.99 rad/s
Capacitive reactance:
XC = 1 / (ωC) = 1 / (376.99 × 100 × 10-6) ≈ 26.53 Ω
Impedance of series RC circuit:
Z = √(R² + XC²) = √(40² + 26.53²) ≈ 48.00 Ω
Rms current:
Irms = Vrms / Z = 110 / 48.00 ≈ 2.29 A
Maximum (peak) current:
I0 = Irms √2 ≈ 2.29 × 1.414 = 3.24 A
Hence, the maximum current in the circuit is approximately 3.24 A.
(b) For a series RC circuit the current leads the voltage by a phase angle Φ where
tan Φ = XC / R = 26.53 / 40 ≈ 0.6633
Φ ≈ arctan(0.6633) ≈ 33.59° = 0.5866 rad
Time lead between current maximum and voltage maximum:
Δt = Φ / ω = 0.5866 / 376.99 ≈ 0.00156 s = 1.56 ms
Hence, the current reaches its maximum about 1.56 ms before the voltage maximum.
Q5. Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
Ans. Capacitance, C = 100 μF = 100 × 10-6 F
Resistance R = 40 Ω
Supply voltage (rms), V = 110 V
Frequency, ν = 12 kHz = 12 × 103 Hz
Angular frequency, ω = 2πν = 2π × 12 × 103 ≈ 75 398 rad/s
Capacitive reactance at high frequency:
XC = 1 / (ωC) ≈ 1 / (75 398 × 100 × 10-6) ≈ 1 / 7.5398 ≈ 0.1326 Ω
Impedance Z ≈ √(R² + XC²) ≈ √(40² + 0.1326²) ≈ 40.0002 Ω
Rms current:
Irms = 110 / 40.0002 ≈ 2.75 A
Peak current I0 ≈ 2.75 × 1.414 ≈ 3.89 A
Phase angle Φ = arctan(XC/R) ≈ arctan(0.1326 / 40) ≈ 0.003315 rad ≈ 0.19° (very small)
Conclusion: At very high frequency XC becomes very small, so the capacitor offers very little impedance and thus behaves like a conductor for high-frequency signals. In a dc circuit after steady state (ω = 0), XC → ∞ and the capacitor behaves like an open circuit, blocking dc current.
Q6. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.
Ans. An inductor (L), a capacitor (C), and a resistor (R) are connected in parallel with each other with:
L = 5.0 H
C = 80 μF = 80 × 10-6 F
R = 40 Ω
Supply (rms) voltage V = 230 V
Resonance angular frequency (from earlier) ω = 50 rad/s
At resonance:
Impedances of the inductor and capacitor branches are equal in magnitude but opposite in phase. Therefore the reactive currents in those two branches are equal in amplitude and opposite in phase, and they cancel in the total current. The only remaining branch current is the resistive branch current which is in phase with the supply voltage. As a result the magnitude of the total current is minimum at resonance (it equals the resistive branch current).
Rms currents in each branch at resonance:
Current through resistor (rms):
IR = V / R = 230 / 40 = 5.75 A
Current through inductor (rms):
IL = V / (ωL) = 230 / (50 × 5.0) = 230 / 250 = 0.92 A
Current through capacitor (rms):
IC = V × ωC = V / (1/(ωC)) = 230 / 250 = 0.92 A
Since IL and IC are equal in magnitude and 180° out of phase, they cancel in the total current.
Hence the total current is minimum (Itotal = IR = 5.75 A), and the rms branch currents are IR = 5.75 A, IL = 0.92 A and IC = 0.92 A.
FAQs on NCERT Solutions: Alternating Current
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