In a driven series RLC circuit, the ideal generator has a peak emf equal to 197

Question

In a driven series RLC circuit, the ideal generator has a peak emf equal to 197 V, the resistance is 59.9 , and the capacitance is 8.02 µF. The inductance can be varied from 7.97 mH to 40.3 mH by the insertion of an iron core in the solenoid. The angular frequency of the generator is 2505 rad/s. If the capacitor voltage is not to exceed 150 V, find the following.

(a) the peak current
A

(b) the ranges of inductance that are safe to use.

-------  mH < L <   ---------mH (lower)
--------- mH < L <   --------mH (higher)

Explanation / Answer

Given,

V = 197 V ; R = 59.9 Ohm ; C = 8.02 uF ; L = 7.97 mH to 40.3 mH ; w = 2505 rad/s ;

a)We know that the current in the circuit will be at peak when Z has least value. Z will be less when we use the least value of inductance.

Ipeak = Vpeak/Z

Z = sqrt (R^2 + (Xl - Xc)^2)

Z = sqrt [59.9^2 + (2505 x 7.97 x 10^-3 - 1/2505 x 8.02 x 10^-6)^2] = 66.91 Ohm

Ipeak = Vpeak/Z

Ipeak = 197/66.91 = 2.94 A

Hence, Ipeak = 2.91 A

b)Since the capacitor voltage should not exceed 150 V

the value of inductance should be such that the threshold the capacitor voltage should not be reached.

drop across the inductor is:

Vl = Xl I = 2505 x 7.97 x 10^-3 x 2.91 = 58.1 V

So the lowest value is 7.97 mH and highest is 40.3 mH

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