A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply.
Question:

A circuit containing a 80 mH inductor and a 60 μF 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 the inductor?

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

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

Solution:

Inductance, L = 80 mH = 80 × 10−3 H

Capacitance, = 60 μF = 60 × 10−6 F

Supply voltage, V = 230 V

Frequency, ν = 50 Hz

Angular frequency, ω = 2πν= 100 π rad/s

Peak voltage, $V_{0}=V \sqrt{2}=230 \sqrt{2} \mathrm{~V}$

(a) Maximum current is given as:

$I_{0}=\frac{V_{0}}{\left(\omega L-\frac{1}{\omega C}\right)}$

$=\frac{230 \sqrt{3}}{\left(100 \pi \times 80 \times 10^{-3}-\frac{1}{100 \pi \times 60 \times 10^{-6}}\right)}$

$=\frac{230 \sqrt{2}}{\left(8 \pi-\frac{1000}{6 \pi}\right)}=-11.63 \mathrm{~A}$

The negative sign appears because $\omega L<\frac{1}{\omega C}$.

Amplitude of maximum current, $\left|I_{0}\right|=11.63 \mathrm{~A}$

Hence, $\mathrm{rms}$ value of current, $I=\frac{I_{0}}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}=-8.22 \mathrm{~A}$

(b) Potential difference across the inductor,

VL= I × ωL

= 8.22 × 100 π × 80 × 10−3

= 206.61 V

Potential difference across the capacitor,

$V_{\mathrm{c}}=I \times \frac{1}{\omega C}$

$=8.22 \times \frac{1}{100 \pi \times 60 \times 10^{-6}}=436.3 \mathrm{~V}$

(c) Average power consumed by the inductor is zero as actual voltage leads the current by $\frac{\pi}{2}$.

(d) Average power consumed by the capacitor is zero as voltage lags current by $\frac{\pi}{2}$.

(e) The total power absorbed (averaged over one cycle) is zero.

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