Q182 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
In series LCR circuit the alternating emf 'e' and the current 'i' are given by the equations $e = e_0 \cos\omega t$, $i = i_0 \sin\left(\omega t + \dfrac{\pi}{6}\right)$. The average power dissipated in the circuit over a one complete cycle of A.C is ($\cos 60 = 1/2$)
- A$\dfrac{e_0 i_0}{2}$
- Correct answerB$\dfrac{e_0 i_0}{4}$
- C$\sqrt{\dfrac{3}{2}} e_0 i_0$
- DZero
Solution
The given equations for alternating emf and current are:
$e = e_0 \cos\omega t = e_0 \sin\left(\omega t + \dfrac{\pi}{2}\right)$
$i = i_0 \sin\left(\omega t + \dfrac{\pi}{6}\right)$
The phase difference $\phi$ between the emf and current is:
$\phi = \dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3} = 60^{\circ}$
The average power dissipated over one complete cycle is given by:
$P_{avg} = e_{rms} i_{rms} \cos\phi$
Substituting the values of $e_{rms}$, $i_{rms}$, and $\cos\phi$:
$P_{avg} = \left(\dfrac{e_0}{\sqrt{2}}\right) \left(\dfrac{i_0}{\sqrt{2}}\right) \cos 60^{\circ}$
$P_{avg} = \dfrac{e_0 i_0}{2} \times \dfrac{1}{2} = \dfrac{e_0 i_0}{4}$
Answer: $\dfrac{e_0 i_0}{4}$
$e = e_0 \cos\omega t = e_0 \sin\left(\omega t + \dfrac{\pi}{2}\right)$
$i = i_0 \sin\left(\omega t + \dfrac{\pi}{6}\right)$
The phase difference $\phi$ between the emf and current is:
$\phi = \dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3} = 60^{\circ}$
The average power dissipated over one complete cycle is given by:
$P_{avg} = e_{rms} i_{rms} \cos\phi$
Substituting the values of $e_{rms}$, $i_{rms}$, and $\cos\phi$:
$P_{avg} = \left(\dfrac{e_0}{\sqrt{2}}\right) \left(\dfrac{i_0}{\sqrt{2}}\right) \cos 60^{\circ}$
$P_{avg} = \dfrac{e_0 i_0}{2} \times \dfrac{1}{2} = \dfrac{e_0 i_0}{4}$
Answer: $\dfrac{e_0 i_0}{4}$
Hence, the correct answer is option: 'B'.