A series LCR circuit consisting of $R = 10\Omega$, $|X_L| = 20\Omega$ and $|X_C| = 20\Omega$, is connected across an a.c. supply of $200\,\text{V}_{\text{rms}}$. The rms voltage across the capacitor is}

Question

Write the principle of working of an ac generator. Draw its labelled diagram and explain its working.

Answer 1

Step 1: Since $|X_L|=|X_C|=20\,\Omega$, the circuit is at resonance and $V_L$ cancels $V_C$, so the full supply appears across $R$: $V_R=200$ V.
Step 2: Current: $I = V_R/R = 200/10 = 20$ A.
Step 3: Capacitor voltage magnitude: $V_C = I\times X_C = 20\times 20 = 400$ V, lagging the current by $90^\circ$.
\[ \boxed{V_C = 400\angle -90^\circ \text{ V}} \]

Answer 2

Working directly with the complex impedance of the series circuit gives the same result through a slightly different route. Taking the inductive reactance as $+j20\,\Omega$ and the capacitive reactance as $-j20\,\Omega$, the total series impedance is \[ Z = R + jX_L - jX_C = 10 + j20 - j20 = 10\,\Omega \] which is purely resistive, confirming resonance directly from the impedance sum rather than from a separate check on the reactance magnitudes.

Taking the supply as the reference phasor $200\angle 0^\circ$ V, the current is \[ I = \frac{200\angle 0^\circ}{10} = 20\angle 0^\circ \text{ A} \] The capacitor's impedance is $-j20\,\Omega = 20\angle -90^\circ\,\Omega$, so the capacitor voltage is \[ V_C = I \times Z_C = 20\angle 0^\circ \times 20\angle -90^\circ = 400\angle -90^\circ \text{ V} \]

$200\angle -90^\circ$ V: This magnitude corresponds to the supply, not to $I\times|Z_C|$, which comes out larger due to the current being magnified relative to what a simple resistive divider would suggest.
$200\angle +90^\circ$ V: Neither the magnitude nor the angle sign match the phasor multiplication above.
$400\angle +90^\circ$ V: The magnitude matches but the angle sign is flipped; multiplying a $0^\circ$ current phasor by a $-90^\circ$ impedance phasor gives $-90^\circ$, not $+90^\circ$.
$400\angle -90^\circ$ V: This matches the direct phasor multiplication $I \times Z_C$ exactly.

Therefore, the correct answer is $400\angle -90^\circ$ V.

Answer 3

Working directly with the complex impedance of the series circuit gives the same result through a slightly different route. Taking the inductive reactance as $+j20\,\Omega$ and the capacitive reactance as $-j20\,\Omega$, the total series impedance is \[ Z = R + jX_L - jX_C = 10 + j20 - j20 = 10\,\Omega \] which is purely resistive, confirming resonance directly from the impedance sum rather than from a separate check on the reactance magnitudes.

Taking the supply as the reference phasor $200\angle 0^\circ$ V, the current is \[ I = \frac{200\angle 0^\circ}{10} = 20\angle 0^\circ \text{ A} \] The capacitor's impedance is $-j20\,\Omega = 20\angle -90^\circ\,\Omega$, so the capacitor voltage is \[ V_C = I \times Z_C = 20\angle 0^\circ \times 20\angle -90^\circ = 400\angle -90^\circ \text{ V} \]

$200\angle -90^\circ$ V: This magnitude corresponds to the supply, not to $I\times|Z_C|$, which comes out larger due to the current being magnified relative to what a simple resistive divider would suggest.
$200\angle +90^\circ$ V: Neither the magnitude nor the angle sign match the phasor multiplication above.
$400\angle +90^\circ$ V: The magnitude matches but the angle sign is flipped; multiplying a $0^\circ$ current phasor by a $-90^\circ$ impedance phasor gives $-90^\circ$, not $+90^\circ$.
$400\angle -90^\circ$ V: This matches the direct phasor multiplication $I \times Z_C$ exactly.

Therefore, the correct answer is $400\angle -90^\circ$ V.

A series LCR circuit consisting of $R = 10\Omega$, $|X_L| = 20\Omega$ and $|X_C| = 20\Omega$, is connected across an a.c. supply of $200\,\text{V}_{\text{rms}}$. The rms voltage across the capacitor is}

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The Correct Option is D

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Approach Solution -2

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