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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 $10\xb0C$. 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.

$\left(A\right){10}^{-15}C\phantom{\rule{0ex}{0ex}}\left(B\right){10}^{-6}C\phantom{\rule{0ex}{0ex}}\left(C\right){10}^{-10}C\phantom{\rule{0ex}{0ex}}\left(D\right){10}^{-8}C$

Considering 10

^{-9}C.

C= 10

^{-11}F, L = 10

^{-5}Henry, and R= 100 ohm

I

_{max}( 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

I

_{max}(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

^{8 }

Q max = Imax /w = 10

^{-8 }

C Ans

â€‹Regards

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