Chapter 7 Alternating Current Solutions
Question - 11 : - Figure 7.21 shows a series LCR circuitconnected to a variable frequency 230 V source. L = 5.0H, C = 80μF, R = 40 Ω
(a) Determinethe source frequency which drives the circuit in resonance.
(b) Obtainthe impedance of the circuit and the amplitude of current at the resonatingfrequency.
(c) Determinethe rms potential drops across the three elements of the circuit. Show that thepotential drop across the LC combination is zero at theresonating frequency.
Answer - 11 : -
Inductance of the inductor, L =5.0 H
Capacitance of thecapacitor, C = 80 μF = 80 × 10−6 F
Resistance of the resistor, R =40 Ω
Potential of the variable voltagesource, V = 230 V
(a) Resonance angularfrequency is given as:Hence, the circuit will come inresonance for a source frequency of 50 rad/s.
(b) Impedance of thecircuit is given by the relation,At resonance,
Amplitude of the current at the resonating frequencyis given as: 
Where,
V0 =Peak voltage
Hence, at resonance, theimpedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.
(c) Rmspotential drop across the inductor,
(VL)rms = I × ωRL
Where,
I =rms current
Potential drop across thecapacitor,
Potential drop across theresistor,
(VR)rms = IR
=
× 40 = 230 V
Potential drop across the LC combination,
At resonance,
∴VLC= 0
Hence, it is proved that thepotential drop across the LC combination is zero at resonatingfrequency.
An LC circuitcontains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit isclosed be t = 0.
(a) What isthe total energy stored initially? Is it conserved during LC oscillations?
(b) What isthe natural frequency of the circuit?
(c) At whattime is the energy stored
(i) completely electrical (i.e.,stored in the capacitor)? (ii) completely magnetic (i.e., stored in theinductor)?
(d) At whattimes is the total energy shared equally between the inductor and thecapacitor?
(e) If aresistor is inserted in the circuit, how much energy is eventually dissipatedas heat?
Answer - 12 : -
Inductance of the inductor, L =20 mH = 20 × 10−3 H
Capacitance of thecapacitor, C = 50 μF = 50 × 10−6 F
Initial charge on thecapacitor, Q = 10 mC = 10 × 10−3 C
(a) Total energy storedinitially in the circuit is given a
Hence, the total energy stored inthe LC circuit will be conserved because there is no resistorconnected in the circuit.
(b)Naturalfrequency of the circuit is given by the relation,
Natural angular frequency,
Hence, the natural frequency ofthe circuit is 103 rad/s.
(c) (i) For time period (T
), total charge on the capacitor at time t,
For energy stored is electrical,we can write Q’ = Q.
Hence,it can be inferred that the energy stored in the capacitor is completelyelectrical at time, t =
(ii) Magneticenergy is the maximum when electrical energy, Q′ is equal to 0.
Hence,it can be inferred that the energy stored in the capacitor is completelymagnetic at time, 
(d) Q1 =Charge on the capacitor when total energy is equally shared between thecapacitor and the inductor at time t.
When total energy is equallyshared between the inductor and capacitor, the energy stored in the capacitor= 
(maximum energy).
Hence, total energy is equally shared between theinductor and the capacity at time,
(e) Ifa resistor is inserted in the circuit, then total initial energy is dissipatedas heat energy in the circuit. The resistance damps out the LC oscillation.
A coil of inductance 0.50 H andresistance 100 Ω is connected to a 240 V, 50 Hz ac supply.
(a) What isthe maximum current in the coil?
(b) What is the time lagbetween the voltage maximum and the current maximum?
Answer - 13 : -
Inductance of the inductor, L =0.50 H
Resistance of the resistor, R =100 Ω
Potential of the supplyvoltage, V = 240 V
Frequency of the supply, ν =50 Hz
(a) Peakvoltage is given as:
Angular frequency of the supply,
ω = 2 πν
= 2π × 50 = 100 π rad/s
Maximum current in the circuit isgiven as:
(b) Equationfor voltage is given as:
V = V0 cos ωt
Equation for current is given as:
I = I0 cos(ωt − Φ)
Where,
Φ = Phase difference betweenvoltage and current
At time, t = 0.
V = V0(voltageis maximum)
Forωt − Φ =0 i.e., at time
I = I0 (currentis maximum)
Hence,the time lag between maximum voltage and maximum current is
Now, phase angle Φisgiven by the relation,
Hence, the time lag betweenmaximum voltage and maximum current is 3.2 ms.
Question - 14 : - Obtain the answers (a) to (b) inExercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10kHz). Hence, explain the statement that at very high frequency, an inductor ina circuit nearly amounts to an open circuit. How does an inductor behave in adc circuit after the steady state?
Answer - 14 : -
Inductance of the inductor, L =0.5 Hz
Resistance of the resistor, R =100 Ω
Potential of the supplyvoltages, V = 240 V
Frequency of the supply,ν = 10 kHz = 104 Hz
Angular frequency, ω =2πν= 2π × 104 rad/s
(a) Peak voltage, 
Maximum current, 
(b) For phasedifferenceΦ, we have the relation:
It can be observed that I0 isvery small in this case. Hence, at high frequencies, the inductor amounts to anopen circuit.
In a dc circuit, after a steadystate is achieved, ω = 0. Hence, inductor L behaves like apure conducting object.
A 100 μF capacitor in series witha 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What isthe maximum current in the circuit?
(b) What isthe time lag between the current maximum and the voltage maximum?
Answer - 15 : -
Capacitance of thecapacitor, C = 100 μF = 100 × 10−6 F
Resistance of the resistor, R =40 Ω
Supply voltage, V =110 V
(a) Frequencyof oscillations, ν= 60 Hz
Angular frequency, 
Fora RC circuit, we have the relation for impedance as:Peak voltage, V0 = 
Maximum current is given as:
(b) In acapacitor circuit, the voltage lags behind the current by a phase angle ofΦ.This angle is given by the relation:
Hence, the time lag betweenmaximum current and maximum voltage is 1.55 ms.
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 thesteady state.
Answer - 16 : -
Capacitance of thecapacitor, C = 100 μF = 100 × 10−6 F
Resistance of the resistor, R =40 Ω
Supply voltage, V =110 V
Frequency of the supply, ν =12 kHz = 12 × 103 Hz
Angular Frequency, ω =2 πν= 2 × π × 12 × 10303
= 24π × 103 rad/s
Peakvoltage, 
Maximum current, 
For an RC circuit,the voltage lags behind the current by a phase angle of Φ givenas:
Hence, Φ tendsto become zero at high frequencies. At a high frequency, capacitor C acts as aconductor.
In a dc circuit, after the steadystate is achieved, ω = 0. Hence, capacitor C amounts to anopen circuit.
Keeping the source frequencyequal to the resonating frequency of the series LCR circuit,if the three elements, L, C and R arearranged in parallel, show that the total current in the parallel LCR circuitis minimum at this frequency. Obtain the current rms value in each branch ofthe circuit for the elements and source specified in Exercise 7.11 for thisfrequency.
Answer - 17 : -
An inductor (L), acapacitor (C), and a resistor (R) is connected in parallel witheach other in a circuit where,
L =5.0 H
C =80 μF = 80 × 10−6 F
R =40 Ω
Potential of the voltagesource, V = 230 V
Impedance (Z) of the givenparallel LCR circuit is given as:
Where,
ω = Angular frequency
Atresonance,
Hence, the magnitude of Z isthe maximum at 50 rad/s. As a result, the total current is minimum.
Rms current flowing throughinductor L is given as:
Rms current flowing throughcapacitor C is given as:
Rms current flowing throughresistor R is given as:
Question - 18 : - A circuit containing a 80 mHinductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply.The resistance of the circuit is negligible.
(a) Obtainthe current amplitude and rms values.
(b) Obtainthe rms values of potential drops across each element.
(c) What isthe average power transferred to the inductor?
(d) What isthe average power transferred to the capacitor?
(e) What isthe total average power absorbed by the circuit? [‘Average’ implies ‘averagedover one cycle’.]
Answer - 18 : -
Inductance, L =80 mH = 80 × 10−3 H
Capacitance, C =60 μF = 60 × 10−6 F
Supply voltage, V =230 V
Frequency, ν =50 Hz
Angular frequency, ω =2πν= 100 π rad/s
Peakvoltage, V0 =
(a) Maximumcurrent is given as:
The negative sign appears because 
Amplitude of maximum current, 
Hence, rms value of current, 
(b) Potentialdifference across the inductor,
VL= I ×ωL
= 8.22 × 100 π × 80 × 10−3
= 206.61 V
Potential difference across thecapacitor,
(c) Averagepower consumed by the inductor is zero as actual voltage leads the current by
(d) Averagepower consumed by the capacitor is zero as voltage lags current by
(e) Thetotal power absorbed (averaged over one cycle) is zero.
Suppose the circuit in Exercise7.18 has a resistance of 15 Ω. Obtain the average power transferred to eachelement of the circuit, and the total power absorbed.
Answer - 19 : -
Average power transferred to theresistor = 788.44 W
Average power transferred to thecapacitor = 0 W
Total power absorbed by thecircuit = 788.44 W
Inductance of inductor, L =80 mH = 80 × 10−3 H
Capacitance of capacitor, C =60 μF = 60 × 10−6 F
Resistance of resistor, R =15 Ω
Potential of voltagesupply, V = 230 V
Frequency of signal, ν =50 Hz
Angular frequency ofsignal, ω = 2πν= 2π × (50) = 100π rad/s
The elements are connected inseries to each other. Hence, impedance of the circuit is given as:
Current flowing in the circuit, 
Average power transferred toresistance is given as:
PR= I2R
= (7.25)2 × 15 =788.44 W
Average power transferred tocapacitor, PC = Average power transferred toinductor, PL = 0
Total power absorbed by thecircuit:
= PR + PC +PL
= 788.44 + 0 + 0 = 788.44 W
Hence, the total power absorbedby the circuit is 788.44 W.
Question - 20 : - A series LCR circuitwith L = 0.12 H, C = 480 nF, R =23 Ω is connected to a 230 V variable frequency supply.
(a) What isthe source frequency for which current amplitude is maximum. Obtain thismaximum value.
(b) What isthe source frequency for which average power absorbed by the circuit is maximum.Obtain the value of this maximum power.
(c) Forwhich frequencies of the source is the power transferred to the circuit halfthe power at resonant frequency? What is the current amplitude at thesefrequencies?
(d) What isthe Q-factor of the given circuit?
Answer - 20 : -
Inductance, L =0.12 H
Capacitance, C =480 nF = 480 × 10−9 F
Resistance, R =23 Ω
Supply voltage, V =230 V
Peak voltage is given as:
V0 = 
= 325.22 V
(a) Current flowing inthe circuit is given by the relation, 
Where,
I0 =maximum at resonance
At resonance, we have
Where,
ωR = Resonanceangular frequency
∴Resonantfrequency, 
And, maximum current 
(b) Maximumaverage power absorbed by the circuit is given as:
Hence, resonant frequency (
) is 
(c) Thepower transferred to the circuit is half the power at resonant frequency.
Frequenciesat which power transferred is half, = 
Where,
Hence, change in frequency, 
∴
And 
Hence, at 648.22 Hz and 678.74 Hzfrequencies, the power transferred is half.
At these frequencies, currentamplitude can be given as:
(d) Q-factorof the given circuit can be obtained using the relation, 
Hence, the Q-factor of the givencircuit is 21.74.