CBSE 2024, Set 1

Question

An inductor, a capacitor and a resistor are connected in series with an ac source $v = v_m \sin \omega t$. Derive an expression for the average power dissipated in the circuit. Also obtain the expression for the resonant frequency of the circuit.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** **Average Power Dissipated:** Let the alternating voltage be $v = v_m \sin \omega t$. In a series LCR circuit, the current leads or lags the voltage by a phase angle $\phi$: $$i = i_m \sin(\omega t + \phi)$$ The instantaneous power is $p = vi$: $$p = (v_m \sin \omega t) 	imes [i_m \sin(\omega t + \phi)]$$ Using the trigonometric identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $$p = \frac{v_m i_m}{2} [\cos(\phi) - \cos(2\omega t + \phi)]$$ The average power over a complete cycle is the average of the two terms. The average value of the time-dependent term $\cos(2\omega t + \phi)$ over a full cycle is zero. Therefore, the average power $P_{av}$ is: $$P_{av} = \frac{v_m i_m}{2} \cos \phi = \left(\frac{v_m}{\sqrt{2}}ight) \left(\frac{i_m}{\sqrt{2}}ight) \cos \phi$$ $$P_{av} = V_{rms} I_{rms} \cos \phi$$ **Resonant Frequency:** Resonance occurs in a series LCR circuit when the capacitive reactance equals the inductive reactance ($X_C = X_L$). $$\frac{1}{\omega C} = \omega L$$ $$\omega^2 = \frac{1}{LC}$$ $$\omega = \frac{1}{\sqrt{LC}}$$ Since $\omega = 2\pi f$: $$f = \frac{1}{2\pi \sqrt{LC}}$$

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Physics > Magnetism Induction And Alternating Current > Alternating Current

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