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A resistor of 100 Omega, inductance of 1...

A resistor of `100 Omega`, inductance of 1 H and a capacitor of capacitance `10.13 xx 10^(-6) F` are in series. This combination is conneceted to an A.C. source of 200 V, 50 Hz. Find the current in the circuit and the P.D. across the resistor.

Text Solution

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Here, `R = 100 Omega, L = 1 H`,
`C = 200 V, v = 50 Hz, I_(v) = ?, V = ?`
`E_(v) = 200 V, v = 50 Hz, I_(v) = ?, V = ?`
`X_(L) = omega L = 2 pi v L = 2 xx 3.14 xx 50 xx 1 = 314 Omega`
`X_(C ) = (1)/(omega C) = (1)/(2 pi v C)`
`= (1)/(2 xx 3.14 xx 50 xx 10.13 xx 10^(-6))`
`= (1000)/(3.14 xx 10.13) = 314.38 Omega ~= 314 Omega`
As `X_(L) = X_(C )`, therefore, circuit is in reasonace.
`I_(v) = (E_(v))/(R ) = (200)/(100) 2 A`
`V = I_(v) xx R = 2 xx 100 = 200 V`
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