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 ?

A 12 Omega resistor and a 0.21 henry inductor are connected in series

1.	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 ?
Answer» `30^@` `40^@` `80^@` `90^@` Solution :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^@`

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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 ?

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