Question

A circuit containing a 100mH inductor, 60uF capacitor and 15ohm resistor in series is
connected to a 230V, 50Hz supply. Calculate impedance and power factor power absorbed.​

Answer 1

(a)

I

o

​

=V

o

​

/(ωL−1/ωC); V

o

​

=

2

​

V

I

o

​

=−11.63⇒magnitude of I

o

​

=11.63

rms current is −11.63/

2

​

=8.22A

(b)

V

C

​

=I/ωC=

2π×50×60×10

−6

8.22

​

=436.3V

V

L

​

=IωL=8.22×2π×50×80×10

−3

=206.5V

(c)

Whatever be the current I in L, actual voltage leads current by π/2. Therefore, average power consumed by L is zero.

(d)

Average power consumed by the capacitor is zero as voltage lags by π/2.

(e)

Since the resistance of circuit is negligible, there is no power absorbed by LC circuit over a cycle. Total average power absorbed is zero.