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A resistor of 100Omega, a pure inductanc...

A resistor of `100Omega`, a pure inductance coil of L=0.5 H and capacitor are in series in a circuit containing an a.c. source of 200 V , 50 Hz . In the circuit , current is ahead of the voltage by `30^@` . Find the value of the capacitance.

Text Solution

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Here `R=100Omega`, L=0.5 H, C= ? , V=200 V , f=50 Hz, `f=30^@`
We have,
`tan phi=(X_C-X_L)/R`
`tan 30^@ =(X_C-X_L)/100 " " because X_L=omegaL` & `X_C=1/(omegaC)`
`0.5xx100=1/(2pifC)-2pifL(because omega=2pif)`
`0.5xx100=1/(2xx3.14xx50C)-2xx3.14xx50xx0.5`
`50=1/(314C)-157`
`50+157=1/(314C)`
`207=1/(314C)`
`C=1/64998=0.00001538`
C=15.38 `muF`
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