Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.21
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 7 Question 21
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Inductance, L = 3.0 H,
Capacitance, C = 27 μF = 27 x 10-6 F,
Resistance, R = 7.4 Ω
At resonance, angular frequency of the source for the given LCR series circuit is given as
ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s
Q-factor of the series:
Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446
To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing ωr, we need to reduce R to half i.e.,
Resistance = R/2 = 7.4/2 = 3.7 Ω