Chapter 7 Alternating Current Solutions
Question - 1 : - A 100 Ω resistor is connected toa 220 V, 50 Hz ac supply.
(a) What is the rms value of current in thecircuit?
(b) What is the net power consumedover a full cycle?
Answer - 1 : -
Resistance of the resistor, R = 100 Ω
Supplyvoltage, V = 220 V
Frequency, ν = 50Hz
(a) The rms value of current in the circuitis given as:
(b) The netpower consumed over a full cycle is given as:
P = VI
=220 × 2.2 = 484 W
Question - 2 : - (a) Thepeak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an accircuit is 10 A. What is the peak current?
Answer - 2 : -
(a) Peakvoltage of the ac supply, V0 = 300 V
Rmsvoltage is given as:(b) Thermsvalue of current is given as:
I =10 A
Now, peak current is given as:
A 44 mH inductor is connected to220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer - 3 : -
Inductance of inductor, L =44 mH = 44 × 10−3 H
Supply voltage, V =220 V
Frequency, ν =50 Hz
Angular frequency, ω= 
Inductive reactance, XL = ω L
Rms value of current is given as:
Hence, the rms value of currentin the circuit is 15.92 A.
Question - 4 : - A 60 μF capacitor is connected toa 110 V, 60 Hz ac supply. Determine the rms value of the current in thecircuit.
Answer - 4 : -
Capacitance of capacitor, C =60 μF = 60 × 10−6 F
Supply voltage, V =110 V
Frequency, ν =60 Hz
Angularfrequency, ω= Capacitive reactance 
Rms value of current is given as:
Hence, the rms value of currentis 2.49 A.
In Exercises 7.3 and 7.4, what isthe net power absorbed by each circuit over a complete cycle. Explain youranswer.
Answer - 5 : -
In the inductive circuit,
Rms value of current, I =15.92 A
Rms value of voltage, V =220 V
Hence, the net power absorbed canbe obtained by the relation,
P = VI cos Φ
Where,
Φ = Phase differencebetween V and I
For a pure inductive circuit, thephase difference between alternating voltage and current is 90° i.e., Φ=90°.
Hence, P = 0i.e., the net power is zero.
In the capacitive circuit,
Rms value of current, I =2.49 A
Rms value of voltage, V =110 V
Hence, the net power absorbed canve obtained as:
P = VI Cos Φ
For a pure capacitive circuit,the phase difference between alternating voltage and current is 90° i.e., Φ=90°.
Hence, P = 0i.e., the net power is zero.
Question - 6 : - Obtain the resonantfrequency ωr of a series LCR circuitwith L = 2.0 H, C = 32 μF and R =10 Ω. What is the Q-value of this circuit?
Answer - 6 : -
Inductance, L =2.0 H
Capacitance, C =32 μF = 32 × 10−6 F
Resistance, R =10 Ω
Resonant frequency is given bythe relation,
Now, Q-value of thecircuit is given as:
Hence, the Q-Value of thiscircuit is 25.
Question - 7 : - A charged 30 μF capacitor isconnected to a 27 mH inductor. What is the angular frequency of free oscillationsof the circuit?
Answer - 7 : -
Capacitance, C =30μF = 30×10−6F
Inductance, L =27 mH = 27 × 10−3 H
Angular frequency is given as:
Hence, the angular frequency offree oscillations of the circuit is 1.11 × 103 rad/s.
Question - 8 : - Suppose the initial charge on thecapacitor in Exercise 7.7 is 6 mC. What is the total energy stored in thecircuit initially? What is the total energy at later time?
Answer - 8 : -
Capacitance of thecapacitor, C = 30 μF = 30×10−6 F
Inductance of the inductor, L =27 mH = 27 × 10−3 H
Charge on the capacitor, Q =6 mC = 6 × 10−3 C
Totalenergy stored in the capacitor can be calculated by the relation,Total energy at a later time willremain the same because energy is shared between the capacitor and theinductor.
A series LCR circuitwith R = 20 Ω, L = 1.5 H and C =35 μF is connected to a variable-frequency 200 V ac supply. When the frequencyof the supply equals the natural frequency of the circuit, what is the averagepower transferred to the circuit in one complete cycle?
Answer - 9 : -
At resonance, the frequency ofthe supply power equals the natural frequency of the given LCR circuit.
Resistance, R =20 Ω
Inductance, L =1.5 H
Capacitance, C =35 μF = 30 × 10−6 F
AC supply voltage to the LCR circuit, V =200 V
Impedance of the circuit is givenby the relation,
At resonance,
Current in the circuit can becalculated as:
Hence, the average powertransferred to the circuit in one complete cycle= VI
= 200 × 10 = 2000 W.
Question - 10 : - A radio can tune over thefrequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). Ifits LC circuit has an effective inductance of 200 μH, whatmust be the range of its variable capacitor?
Answer - 10 : -
The range of frequency (ν) of aradio is 800 kHz to 1200 kHz.
Lower tuning frequency, ν1 =800 kHz = 800 × 103 Hz
Upper tuning frequency, ν2 =1200 kHz = 1200× 103 Hz
Effective inductance ofcircuit L = 200 μH = 200 × 10−6 H
Capacitanceof variable capacitor for ν1 is given as:C1
Where,
ω1 = Angularfrequency for capacitor C1
Capacitance of variable capacitorfor ν2,
C2 
Where,
ω2 = Angularfrequency for capacitor C2
Hence, the range of the variablecapacitor is from 88.04 pF to 198.1 pF.