In a circuit, a capacitor of \(40\,\mu\mathrm{F}\) and inductor of \(60\,\mathrm{mH}\) are connected in series to a \(200\,\mathrm{V}\) and \(50\,\mathrm{Hz}\) power supply. Suppose the resistance of the circuit is negligible, then the rms value of current is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3.29\,\mathrm{A}\)
With negligible resistance, the impedance of a series LC circuit is the difference of the two reactances.
\( \displaystyle X_L = 2\pi fL = 2\pi \times 50\,\mathrm{Hz} \times 60\times10^{-3}\,\mathrm{H} \approx 18.85\,\Omega \)
\( \displaystyle X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\,\mathrm{Hz} \times 40\times10^{-6}\,\mathrm{F}} \approx 79.58\,\Omega \)
\( \displaystyle Z = |X_C - X_L| = 79.58\,\Omega - 18.85\,\Omega = 60.73\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z} = \frac{200\,\mathrm{V}}{60.73\,\Omega} \)
\( \displaystyle I_{\mathrm{rms}} \approx 3.29\,\mathrm{A} \)
The 200 V of the supply is its rms value, so this is the rms current.