Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.20
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 7 Question 20
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Inductance, L = 0.12 H
Capacitance, C = 480 nF = 480 x 10⁻9 F
Resistance, R = 23 Ω
Supply voltage, V = 230 V
Peak voltage is given as: Vo =√2 x 230 = 325.22 V
Ans (a).
Current flowing in the circuit is given by the relation,
I0= V0/√[(R2 + (ωL- 1/ωC)2]
Where, Io = maximum at resonance
At resonance, we have
ωR– 1/ωRC =0
Where, ωR = Resonance angular frequency
Where , ωR = Resonance angular frequency
Therefore ωR = 1/√(LC) = 1/ √ (0.12 x 480 x 10⁻9 ) = 4166.67 rad/s
Resonant frequency , νR = ωR/2π = 4166.67 /(2 x 3.14) = 663.48 Hz
And ,maximum current, (I0)Max = V0/R = 325.22/23 = 14.14 A
Ans (b).
Maximum average power absorbed by the circuit is given as:
(Pav)Max= 1/2 x (I0)²MaxR
= 1/2 x (14.14)² x 23 = 2299.3 W
Hence, resonant frequency is 663.48 Hz.
Ans (c).
The power transferred to the circuit is half the power at resonant frequency.
Frequencies at which power transferred is half,
= ωR ± Δω
= 2π (νR ± Δν)
Where,
Δω=R/2L= 23/(2 x 0.12)= 95.83 rad/s
Hence, change in frequency,
Δν = 1/2π x Δω =95.83 /2π = 15.26 Hz
Therefore
νR +Δν = 663.48 + 15.26 = 678.74 Hz
and νR – Δν = 663.48 – 15.26 =648.22 Hz
Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half. At these frequencies, current amplitude can be given as:
I’ = 1/√ 2 x (I0)Max
= 14.14/√ 2 = 10A
Ans (d).
Q-factor of the given circuit can be obtained using the relation,
Q = ωR L/R = 4166.67 x 0.12 /23 = 21.74
Hence, the Q-factor of the given circuit is 21.74.