In a series LR circuit, the voltage drop across inductor is 8 volt and across resistor is 6 volt. Then voltage applied and power factor of circuit respectively are:

Options

A. Volatage of the source will be leading current in the ciruit
B. Volatage drop across each element will be less the appliced voltage
C. Power factor of circuit will be 4/3
D. None of these

Answer

D

Explanation

Since cos `theta = (R)/(Z) = (IR)/(IZ) = (8)/(10) = (4)/(5) `
(cos `theta` can never be greater than 1)
Also , `IX_(C) gt IX_(L) implies X_(C) gt X_(L)`
Current will be leading
In a LCR circuit
`V = sqrt((V_(L) - V_(C))^(2) + V_(R)^(2)) = sqrt((6 - 12)^(2) + 8^(2))`
V = 10 , which is less than voltage drop across capacitor .