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 . 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.
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 = 108
Q max = Imax /w = 10 -8
C Ans
​Regards
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 = 108
Q max = Imax /w = 10 -8
C Ans
​Regards