Question

A circuit containing a 80 mH inductor and a $60\mu F$ capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.(a) Obtain the current amplitude and rms values.

(b) Obtain the rms values of potential drops across each element.

(c) What is the average power transferred to

(d) What is the average power transferred to
the capacitor?

(e) What is the total average power absorbed by the circuit? [Average implies average value over one cycle.]

Answer 1

(a)

$I_o=V_o/(\omega L-1/\omega C);V_o=\sqrt{2}V$

$I_o=-11.63\Rightarrow$magnitude of $I_o=11.63$

rms current is $-11.63/\sqrt{2}=8.22A$

(b)

$V_C=I/\omega C=\frac{8.22}{2\pi \times 50\times 60\times {10}^{-6}}=436.3V$

$V_L=I\omega L=8.22\times 2\pi \times 50\times 80\times {10}^{-3}=206.5V$

(c)

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

(d)

Average power consumed by the capacitor is zero as voltage lags by $\pi /2$.

(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.