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

Solution:

Inductance, L = 80 mH = 80 × 10−3 H

Capacitance, C = 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.