Verified Solution • 2025–26 Session moderate

QJEEM24Phbef5: JEE Main 2024 Physics - Class 12 JEEM24Phbef5 Solution: In an a.c. circuit, voltage and current are given by: and…

In an a.c. circuit, voltage and current are given by:

V=100sin⁡(100t)VV=100 \sin (100 t) V and I=100sin⁡(100t+π3)mAI=100 \sin \left(100 t+\frac{\pi}{3}\right) \mathrm{mA} respectively.

The average power dissipated in one cycle is:

Multiple Choice Options

  • A 5 W
  • B 25 W
  • C 2.5 W
  • D 10 W

Step-by-Step Verified Solution

amp;Pavg =Vrms Irmscos⁡(Δϕ)amp;=1002×100×10−32×cos⁡(π3)amp;=1042×12×10−3amp;=104=2.5 W\begin{aligned}& P_{\text {avg }}=V_{\text {rms }} I_{r m s} \cos (\Delta \phi) \\& =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos \left(\frac{\pi}{3}\right) \\& =\frac{10^4}{2} \times \frac{1}{2} \times 10^{-3} \\& =\frac{10}{4}=2.5 \mathrm{~W}\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring