A circuit consists of a `2 mu F` capacitor and a resistor of `1 kOmega`. An a.c. source of 12 V, 50 hz is connected across the circuit. Calculate (i) current flowing (ii) voltage across capacitor (iii) phase angle between voltage and current (iv) average power supplied.

Explanation

Here, `C = 2 mu F = 2 xx 10^(-6) F, R = 1 kOmega = 10^(3) Omega`
`E_(v) = 12 V, v = 50 hz`
`X_(C ) = (1)/(omega C) = (1)/(2 pi v C) = (1)/(2 xx 3.14 xx 50 xx 2 xx 10^(-6))`
` = 1592.36 Omega`
`Z = sqrt(R^(2) + X_(C )^(2)) = sqrt((10^(3))^(2) + (1592.36)^(2))`
`1.88 xx 10^(3) Omega`
`I_(v) = (E_(v))/(Z) = (12)/(1.88 xx 10^(3)) = 6.4 xx 10^(-3) A`
Voltage across capacitor
`= I_(v) X_(C ) = 6.4 xx 10^(-3) xx 1592.36 = 10.2 V`
`tan phi = (X_(C )) /(R ) = (1592.36)/(10^(3)) = 1.592`
`= phi = 50^(@)`
Power supplied `I_(v)^(2) R = (6.4 xx 10^(-3))^(2) xx 10^(3)`
`= 40.96 xx 10^(-3) W = 0.041 W`