1.

A resistance of 300 Omega and inductance of 1//pi henry are connected in series to an alternating voltage of 20 V and 200 Hz frequency. The phase angle between the voltage and current is

Answer»

`tan^(-1)((3)/(4))`
`tan^(-1)((3)/(2))`
`tan^(-1)((4)/(3))`
`tan^(-1)((2)/(3))`

Solution :Phase angle `= phi`
`THEREFORE tan phi = (L omega)/(R )=(2pi fL)/(R )=(2pi xx 200)/(300 xx pi)=(4)/(3) rArr phi = tan^(-1)((4)/(3))`


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