In an a.c. series LCR circuit, the applied voltage is \( V = (100\sqrt{2})\sin(\omega t + \pi/4) \) volt. Given that, \( R = 30\,\Omega \), \( X_L = 50\,\Omega \) and \( X_C = 10\,\Omega \). The value of power dissipated in the circuit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 120 W
In a series LCR circuit the average power is dissipated only in the resistor, \( P = I_{\mathrm{rms}}^{2}R \). The current follows from the impedance; the phase \( \pi/4 \) of the source does not matter.
\( \displaystyle V_{\mathrm{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\,\mathrm{V} \)
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{30^{2} + (50 - 10)^{2}} = \sqrt{900 + 1600} = 50\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z} = \frac{100}{50} = 2\,\mathrm{A} \)
\( \displaystyle P = I_{\mathrm{rms}}^{2}R = 2^{2} \times 30 = 120\,\mathrm{W} \)
The same value follows from \( P = V_{\mathrm{rms}} I_{\mathrm{rms}} \cos\phi \) with power factor \( \cos\phi = R/Z = 0.6 \): \( 100 \times 2 \times 0.6 = 120\,\mathrm{W} \).