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Exercises
Question 7.1:
A 100 ฮฉ 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 7.1:
Resistance of the resistor, R = 100 ฮฉ
Supply voltage, V = 220 V
Frequency, ฮฝ = 50 Hz
(a) The rms value of current in the circuit is given as ๐ผ =
(b) The net power consumed over a full cycle is given as:
P = VI = 220 ร 2.2 = 484 W
Question 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 7.2: (a) Peak voltage of the ac supply, V 0 = 300 V rms voltage is given as:
๐ =
(b) The rms value of current is given as: I = 10 A Now, peak current is given as: ๐ผ๐ = โ2๐ผ = โ2 ร 10 = 14.1 ๐ด
Question 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 7.3:
Inductance of inductor, L = 44 mH = 44 ร 10โ^3 H
Supply voltage, V = 220 V
Frequency, ฮฝ = 50 Hz Angular frequency, ฯ = 2๐ฮฝ Inductive reactance, XL = ฯ L = 2๐ฮฝL = 2๐ ร 50 ร 44 ร 10โ3^ ฮฉ
rms value of current is given as:
๐๐ฟ^ =^
2๐ ร 50 ร 44 ร 10โ3^ = 15.92 ๐ด
Hence, the rms value of current in the circuit is 15.92 A.
Question 7.4:
A 60 ฮผF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the
rms value of the current in the circuit.
Answer 7.4:
Capacitance of capacitor, C = 60 ฮผF = 60 ร 10โ^6 F
For a pure inductive circuit, the phase difference between alternating voltage
and current is 90ยฐ i.e., ๐ท= 90ยฐ.
Hence, P = 0 i.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 can be obtained as:
๐ = ๐๐ผ ๐ถ๐๐ ๐ท
For a pure capacitive circuit, the phase difference between alternating
voltage and current is 90ยฐ i.e., ๐ท = 90ยฐ.
Hence, P = 0 i.e., the net power is zero.
Question 7.6:
Obtain the resonant frequency ฯr of a series LCR circuit with L = 2.0 H,
C = 32 ฮผF and R = 10 ฮฉ. What is the Q-value of this circuit?
Answer 7.6:
Inductance, L = 2.0 H
Capacitance, C = 32 ฮผF = 32 ร 10โ^6 F
Resistance, R = 10 ฮฉ
Resonant frequency is given by the relation,
โ2 ร 32 ร 10โ^
8 ร 10โ3^ = 125 ๐๐๐/๐ .
Now, Q-value of the circuit is given as:
32 ร 10โ^
10 ร 4 ร 10โ^
Hence, the Q-Value of this circuit is 25.
Question 7.7:
A charged 30 ฮผF capacitor is connected to a 27 mH inductor. What is the
angular frequency of free oscillations of the circuit?
Answer 7.7:
Capacitance, C = 30ฮผF = 30ร10โ^6 F Inductance, L = 27 mH = 27 ร 10โ^3 H
Angular frequency is given as:
๐๐ =
โ27 ร 10โ3^ ร 30 ร 10โ^
9 ร 10โ4^ = 1.11 ร 10
Hence, the angular frequency of free oscillations of the circuit is 1.11 ร 10^3 rad/s.
Question 7.8:
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 total energy at later
time?
Answer 7.8:
Capacitance of the capacitor, C = 30 ฮผF = 30ร10โ^6 F
Current in the circuit can be calculated as:
๐ผ =
Hence, the average power transferred to the circuit in one complete cycle:
VI = 200 ร 10 = 2000 W.
Question 7.10:
A radio can tune over the frequency range of a portion of MW broadcast
band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of
200 ฮผH, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations
of the LC circuit should be equal to the frequency of the radio wave.]
Answer 7.10:
The range of frequency (ฮฝ) of a radio is 800 kHz to 1200 kHz.
Lower tuning frequency, ฮฝ 1 = 800 kHz = 800 ร 10^3 Hz
Upper tuning frequency, ฮฝ 2 = 1200 kHz = 1200 ร 10^3 Hz
Effective inductance of circuit L = 200 ฮผH = 200 ร 10โ^6 H
Capacitance of variable capacitor for ฮฝ 1 is given as:
๐ถ 1 =
Where,
ฯ 1 = Angular frequency for capacitor C 1 = 2๐๐ 1 = 2๐ ร 800 ร 10^3 ๐๐๐/๐
(2๐ ร 800 ร 10^3 )^2 ร 200 ร 10โ
= 1.9809 ร 10โ10^ ๐น = 198 ๐๐น
Capacitance of variable capacitor for ฮฝ 2 is given as:
๐ถ 2 =
Where,
ฯ 2 = Angular frequency for capacitor C 2 = 2๐๐ 2 = 2๐ ร 1200 ร 10^3 ๐๐๐/๐
(2๐ ร 1200 ร 10^3 )^2 ร 200 ร 10โ
= 0.8804 ร 10โ10^ ๐น = 88 ๐๐น
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.
Question 7.11:
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80ฮผF, R = 40 ฮฉ
Where,
๐๐ = Peak voltage = โ2 ๐
โด ๐ผ๐ =
โ2 ร 230
Hence, at resonance, the impedance of the circuit is 40 ฮฉ and the amplitude
of the current is 8.13 A.
(c) rms potential drop across the inductor, (VL)rms = I ร ฯrL
Where,
๐ผrms =
โด (๐๐ฟ)rms =
ร 50 ร 5 = 1437.5 ๐
Potential drop across the capacitor:
โด (๐๐ถ )rms = ๐ผ ร
4 ร^
50 ร 80 ร 10โ6^ = 1437.5 ๐
Potential drop across the resistor:
(๐๐
)rms = ๐ผ๐
=
4 ร 40 = 230 ๐
Potential drop across the LC combination: ๐๐ฟ๐ถ = ๐ผ(๐๐ฟ โ ๐๐ถ ) At resonance, ๐๐ฟ = ๐๐ถ โ ๐๐ฟ๐ถ = 0 Hence, it is proved that the potential drop across the LC combination is zero
at resonating frequency.
