13) In a series LCR circuit, C = 10-11 , Farad, L = 10-5 Henry and R = 100 ohm, when a constant D.C voltage E is applied to the circuit, the capacitor acquires a charge 10 ° C . The D.C source is replaced by a sinusoidal voltage source in which the peak voltage E. At resonance the peak value of the charge acquired by the capacitor will be.
( A )   10 - 15 C ( B )   10 - 6 C ( C )   10 - 10 C ( D )   10 - 8 C

Dear Student,


Considering 10-9C.

C= 10-11F, L = 10-5Henry, and R= 100 ohm

Imax ( wL - 1/wC + R ) = V max   ..............(1)
wL = inductive reactance
1/wC = capacitive reactance
R is resistance

at resonance :    wl = 1/wC
so equation(1) becomes
Imax(R) = Vmax
I max = Eo/R ......(1)

Calculation of Eo
initially the circuit is connected with constant battery source:
at steady state potential across capacitor becomes = E = Eo
we know that 
q = C E   ( Given at steady state q= 10-9 C)
put values and get E as
E = Eo = 10-9/10-11  = 100 volt.............(2)
Put value of Eo in 1
Get I max = 1 amp
now at resonance w = (1/LC)1/2 = 10 
Q max = Imax /w = 10 -8 

C Ans

​Regards

 

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