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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Alternating Current

JEE Main 2018 · 8 April · Physics · Alternating Current

In an A.C. circuit, the instantaneous e.m.f. and current are given...

In an A.C.\text{A.C.} circuit, the instantaneous e.m.f. and current are given by, E = 100\sin 30t, I = 20\sin \left(30t -π4\dfrac{\pi}{4}\right).In. Inone cycle of A.C.\text{A.C.}, the average power consumed by the circuit (in watt) and the watt-less current (in ampere) are, respectively:In an A.C. circuit, the instantaneous e.m.f. and current are given by, E = 100 30t, I = 20 (30t - (π)/(4) ) . In one

Options

  1. A50,050, 0
  2. B50,1050, 10
  3. C10002,10\dfrac{1000}{\sqrt{2}}, 10
  4. D502,0\dfrac{50}{\sqrt{2}}, 0

Correct answer

C. 10002,10\dfrac{1000}{\sqrt{2}}, 10

Step-by-step solution

Instantaneous emf and current is given by, E=100sin⁡30tE = 100\sin 30t, I=20sin⁡(30t−π4)I = 20\sin \left(30t - \dfrac{\pi}{4}\right).

Phase difference between emf and current is

ϕ=30t−(30t−π4)=π4ϕ = 30t - \left(30t - \dfrac{\pi}{4}\right) = \dfrac{\pi}{4}

In an A.C. circuit, the instantaneous e.m.f. and current are given by, E = 100 30t, I = 20 (30t - (π)/(4) ) . In one

Average power is given by,

Pav=ErmsIrmscosϕ{P}_{av} = {E}_{rms}{I}_{rms}cosϕ
=1002×202×12= \dfrac{100}{\sqrt{2}} \times \dfrac{20}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}}=10002= \dfrac{1000}{\sqrt{2}}



Watt-less current is given by,

IWL=Irmssinϕ{I}_{WL} = {I}_{rms}sinϕ=Io√2sin⁡45°= \dfrac{{I}_{o}}{√ 2}\sin 45 °⇒IWL=202.12=10A⇒ {I}_{WL} = \dfrac{20}{\sqrt{2}} . \dfrac{1}{\sqrt{2}} = 10A.

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