1.

A coil has inductance of 0.4 H and resistance of 8Omega . It is connected to an AC source with peak emf 4 V and frequency (30)/(pi) Hz. The average power. dissipated in the circuit is

Answer»

1W
0.5
0.3 W
0.1 W

Solution :Average power dissipated is
`P_(avg) = V_(rms) I_(rms) COS phi`
` = (V_(MAX) )/(sqrt2) xx ( (V_(max) )/(sqrt2) ) xx 1/Z xx R/Z`
` = (V_(max)^2)/(2) xx (R )/(cancelZ^2) = (V_(max)^2) xx (R )/((sqrt(X_L^2 + R^2) )`
` ((4)^2)/(2) xx (8))/(sqrt(0.4 xx 60)^2 + 8^2)^2)`
` = (16 xx 8)/(2 xx (24^2 + 8^2) ) = 64/640= 1/10 = 0.1 W`


Discussion

No Comment Found

Related InterviewSolutions