An LCR series circuit, with \( L = \displaystyle\frac{2}{\pi}\,\mathrm{H} \), \( R = 100\,\Omega \) and \( C = \displaystyle\frac{100}{\pi}\,\mu\mathrm{F} \) is connected across a source of 200 V, 50 Hz as shown. The peak potential difference between A and B is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 200\sqrt{5}\,\mathrm{V} \)
In a series LCR circuit the same current flows through every element. The voltage between A and B is across the inductor and the resistor together, so its peak value is the peak current times the impedance of the L-R part. The 200 V of the source is its rms value.
\( \displaystyle X_L = 2\pi f L = 2\pi \times 50 \times \frac{2}{\pi} = 200\,\Omega \)
\( \displaystyle X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{100}{\pi}\times10^{-6}} = \frac{1}{0.01} = 100\,\Omega \)
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{100^{2} + 100^{2}} = 100\sqrt{2}\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z} = \frac{200}{100\sqrt{2}} = \sqrt{2}\,\mathrm{A}, \qquad I_0 = \sqrt{2}\,I_{\mathrm{rms}} = 2\,\mathrm{A} \)
Across L and R together:
\( \displaystyle Z_{AB} = \sqrt{R^{2} + X_L^{2}} = \sqrt{100^{2} + 200^{2}} = 100\sqrt{5}\,\Omega \)
\( \displaystyle V_{AB,0} = I_0\,Z_{AB} = 2 \times 100\sqrt{5} = 200\sqrt{5}\,\mathrm{V} \)