Q194 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
In an AC circuit, E and I are given by $E = 150\sin(150t)$ V and $I = 150\sin\left(150t + \dfrac{\pi}{3}\right)$ A. The power dissipated in the circuit is
$\{\cos(60)^\circ = 1/2\}$
$\{\cos(60)^\circ = 1/2\}$
- A$106\ \text{W}$
- B$150\ \text{V}$
- Correct answerC$5625\ \text{W}$
- Dzero
Solution
The given equations are $E = 150\sin(150t)$ and $I = 150\sin\left(150t + \dfrac{\pi}{3}\right)$.
The peak voltage $E_0 = 150$ V and the peak current $I_0 = 150$ A.
The phase difference between voltage and current is $\phi = \dfrac{\pi}{3} = 60^\circ$.
The average power dissipated in the circuit is given by $P = E_{rms} I_{rms} \cos\phi$.
$P = \dfrac{E_0}{\sqrt{2}} \dfrac{I_0}{\sqrt{2}} \cos\phi = \dfrac{E_0 I_0}{2} \cos\phi$
Substituting the values:
$P = \dfrac{150 \times 150}{2} \cos(60^\circ)$
$P = \dfrac{22500}{2} \times \dfrac{1}{2} = \dfrac{22500}{4} = 5625$ W.
Answer: $5625\ \text{W}$
The peak voltage $E_0 = 150$ V and the peak current $I_0 = 150$ A.
The phase difference between voltage and current is $\phi = \dfrac{\pi}{3} = 60^\circ$.
The average power dissipated in the circuit is given by $P = E_{rms} I_{rms} \cos\phi$.
$P = \dfrac{E_0}{\sqrt{2}} \dfrac{I_0}{\sqrt{2}} \cos\phi = \dfrac{E_0 I_0}{2} \cos\phi$
Substituting the values:
$P = \dfrac{150 \times 150}{2} \cos(60^\circ)$
$P = \dfrac{22500}{2} \times \dfrac{1}{2} = \dfrac{22500}{4} = 5625$ W.
Answer: $5625\ \text{W}$
Hence, the correct answer is option: 'C'.