1.

An L-C-R series circuit with R = 100Omega is connected to a 200 V, 50 Hz A.C. source when only the capacitance is removed, the current lags the voltage by 60^@. When only the inductance is removed, the current leads the voltage by 60^@. Find the current in the circuit.

Answer»

`4 A`
`2 A`
`6 A`
`8 A`

Solution :Removing capacitor, L - R circuit is prepared where phase difference `delta=60^@`
`therefore` tan `60^@ =X_L/R=sqrt3 rArr X_L=sqrt3R`….(1)
Removing inductor C-R circuit is prepared where phase difference `|delta|=60^@`
`therefore tan 60^@ = X_C/R = sqrt3 rArr X_C=sqrt3R`....(2)
From (1) and (2) `X_L = X_C rArr omegaL=1/(omegaC)`
means RESONANCE circuit is prepared
`therefore Z=R rArr I_(rms) = V_(rms)/R`
`=200/100`=2A


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