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A 400 Omega resistor, a 3 H inductor and...

A 400 `Omega` resistor, a 3 H inductor and a 5 `muF` capacitor are connected in series to a 220 V, 50 Hz ac source. Calculate the :
(1) Impedance of the ciruit.
(ii) Current flowing through the circuit.

Text Solution

Verified by Experts

`R= 400Omega, L= 3H, C= 5mu F= 5 xx 10^(-6)F V= 220V, f= 50Hz`
Impedance `Z= sqrt(R^(2) + (X_(L)- X_(C ))^(2))`
`X_(L)= 2pi f L = 2pi xx 50 xx 3= 300pi Omega`
`X_(C )= (1)/(2pi fC)= (1)/(2pi xx 50 xx 5 xx 10^(-6))= (2)/(pi) xx 10^(3) Omega`
`Z= sqrt(R^(2) + (X_(L)-X_(C ))^(2))= sqrt(400^(2) + (300pi - (2)/(pi) xx 10^(3))^(2))`
`=503.54Omega`
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