In the given circuit \( L = \displaystyle\frac{20}{\pi}\,\mathrm{H} \), \( R = 2\,\mathrm{k}\Omega \) and \( C = \displaystyle\frac{5}{\pi}\,\mu\mathrm{F} \) are in series across a 220 V, 50 Hz ac source. The peak voltage drop between A and B is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 440 V
Points A and B enclose only R and C, so the voltage between them is the voltage across the R-C part. First find the current in the whole series circuit, then the impedance between A and B.
\( \displaystyle \omega = 2\pi f = 2\pi \times 50 = 100\pi\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle X_L = \omega L = 100\pi \times \frac{20}{\pi} = 2000\,\Omega \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{5}{\pi}\times10^{-6}} = 2000\,\Omega \)
Since \( X_L = X_C \), the circuit is at resonance, the total impedance is \( Z = R = 2000\,\Omega \), and
\( \displaystyle I_{\mathrm{rms}} = \frac{220}{2000} = 0.11\,\mathrm{A} \)
The impedance of the part between A and B (R in series with C) is
\( \displaystyle Z_{AB} = \sqrt{R^{2} + X_C^{2}} = \sqrt{2000^{2} + 2000^{2}} = 2000\sqrt{2}\,\Omega \)
\( \displaystyle V_{AB,\mathrm{rms}} = I_{\mathrm{rms}}\,Z_{AB} = 0.11 \times 2000\sqrt{2} = 220\sqrt{2}\,\mathrm{V} \)
The peak value is \( \sqrt{2} \) times the rms value:
\( \displaystyle V_{AB,0} = \sqrt{2} \times 220\sqrt{2} = 440\,\mathrm{V} \)