A 12 `Omega` resistor and a 0.21 henry inductor are connected in series to an A.C. source operating at 20 volt, 50 cycle/second. What is the phase angle between the current and the source voltage ?

Options

A. `30^@`
B. `40^@`
C. `80^@`
D. `90^@`

Answer

C

Explanation

Here R - L are connected in series so phase difference between V and I is
`tan delta =(omegaL)/R=(2pif)/R=(2xx3.14xx50xx0.21)/12`
`therefore tan delta`= 5.495
`therefore delta=79^@ 42.` (approximate )
`therefore delta approx 80^@`