Question

An inductor L, capacitor of `20 mu F` and a resistor of `10 Omega` are connected in series with 220 V, 50 Hz a.c. supply. If current is in phase with the voltage, calculate the inductance. What is the value of current in the circuit ?

Answer 1

Here, `C = 20 mu F = 20 xx 10^(-6) F, R = 10 Omega`
`E_(v) = 220 V, v = 50 Hz, L = ? I_(v) = ?`
As current is in phase with the voltage,
`X_(L) = X_(C )`
`omega L = (1)/(omega C)`,
`L = (1)/(omega^(2) C ) = (1)/((2 pi v )^(2) C)`
`L = (1)/((2 pi xx 50)^(2) xx 20 xx 10^(-6)) = 0.506 H`
` Z = sqrt(R^(2) + (x_(L) - X_(C ))^(2)) = R = 10 Omega`
`I_(v) = (E_(v))/(Z) = (220)/(10) = 22 A`