CBSE Class 12 Physics 2024: question 25

Hints

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Step-by-step solution

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1. In a series LCR circuit with $v=v_m\sin \omega t$, the current is $i=i_m\sin (\omega t+\varphi )$, where $\tan \varphi =\frac{X_C-X_L}{R}$.

   ½ mark

2. Instantaneous power: $p=vi=v_m{i}_m\sin \omega t\sin (\omega t+\varphi )=\frac{v_m{i}_m}{2}[\cos \varphi -\cos (2\omega t+\varphi )]$.

   ½ mark

3. Over a complete cycle, $\cos \varphi$ is constant. The time-dependent term $\cos (2\omega t+\varphi )$ averages to zero.

   ½ mark

4. Hence $\bar{P}=\frac{v_m{i}_m}{2}\cos \varphi =\frac{v_m}{\sqrt{2}}\frac{i_m}{\sqrt{2}}\cos \varphi =VI\cos \varphi =I^{2}Z\cos \varphi$, with power factor $\cos \varphi =R/Z$.

   ½ mark

5. At resonance the impedance is minimum and the current is maximum, which happens when $X_L=X_C$, i.e. $\omega L=\frac{1}{\omega C}$.

   ½ mark

6. ${\omega }_0=\frac{1}{\sqrt{LC}}$, so the resonant frequency is ${\nu }_0=\frac{1}{2\pi \sqrt{LC}}$.

   ½ mark

Answer: $\bar{P}=V_{rms}{I}_{rms}\cos \varphi$; ${\nu }_0=\frac{1}{2\pi \sqrt{LC}}$

Revise the chapter

• Chapter: Alternating Current (notes, formulas, where marks are lost and every past question)
• Topic: Series LCR Circuit and Resonance