Alternating Current

JEE Main level, single correct, medium

16 of 190

Question 16 of 190

In an a.c. circuit, voltage and current are given by: V=100sin⁡(100t)VV=100\sin (100t)V and I=100sin⁡(100t+π3)mAI=100\sin \left(100t+\dfrac{\pi }{3}\right)mA respectively. The average power dissipated in one cycle is :

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Answer: (C) 2.5W2.5W

Standard expression of voltage is given by, V=V0sin⁡(ωt)V={V}_{0}\sin \left(\omega t\right) and for current is given by, I=I0sin⁡(ωt+ϕ)I={I}_{0}\sin \left(\omega t+\phi \right). Clearly, ϕ\phi is the phase difference here. From the given expression, ϕ=π3\phi =\dfrac{\pi }{3}.
Average power dissipated in one cycle is given by,
Pavg=VrmsIrmscos⁡(ϕ){P}_{avg}={V}_{rms}{I}_{rms}\cos (\phi )
=1002×100×10−32×cos⁡(π3)=\dfrac{100}{\sqrt{2}}\times \dfrac{100\times {10}^{-3}}{\sqrt{2}}\times \cos \left(\dfrac{\pi }{3}\right)
=1042×10−3×12=\dfrac{{10}^{4}}{2}\times {10}^{-3}\times \dfrac{1}{2}
=104=2.5W=\dfrac{10}{4}=2.5W

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