A seriesLCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is `pi/3`. If instead C is removed from the circuit phase difference is agin `pi/3` between current and voltage. Power factor of circuit is:

Options

A. Zero
B. 0.5
C. 1
D. `-1.0`

Answer

C

Explanation

When L is removed
`tan phi = (|X_(C)|)/(R) rArr tan "" (pi)/(3) = (X_(C))/( R)" "` . . . (i)
When C is removed
`tan phi = (|X_(L)|)/(R) rArr tan "" (pi)/(3) = m(X_(L))/(R)" "` . . . (ii)
From (i) and (ii), `X_(L) = X_(C)`
Since, `X_(L) = X_(C)` the circuit is in resonance Z = R
Power factor `= cos phi = (R)/(Z) = 1 `