Question

For the circuit shown in Fig. the emf of the generator is E. The current through the inductor is 1.6 A. While the current through the condenser is 0.4 A. Then

A. Current drawn from the generator is `I=2A`
B. Current drawn from the generator is `I=1.2A`
C. `omega=(1)/(2sqrt(LC))`
D. `omega=(1)/(4sqrt(LC))`

Answer 1

Correct Answer - B::C
`I=I_(C)+I_(L)`
`implies I=I=(E_(0))/(X_(C))` `sin(omega t+pi//2)+(E_0)/(X_L) sin (omegat-pi//2)`
`=[(E_0)/(X_C)-(E_0)/(X_L)] cos omega t`
to find, `I_(v)=|(1)/(sqrt(2)[(E_0)/(X_C)-(E_0)/(X_L)]|`...(i)
Given `I_(Cv)=(E_0)/(sqrt(2)X_(C))=0.4A`, ...(ii) `I_(Lv)=(E_0)/(sqrt(2)X_(L))=1.6 A` ...(iii)
From equations (i) (ii) and (iii) `I_(v)=1.2A`
Dividing (ii) by (iii)
`(X_L)/(X_C)=1/4 implies (omegaL)/(1//omegaC)=1/4 implies omega^(2)LC= 1/4 implies omega=(1)/(2sqt(LC))`.