### JEE Main On 08 April 2018 Question 27

##### Question: In an a.c. circuit, the instantaneous e.m.f. and current are given by $ \text{e=100 sin 30 t} $ $ \text{i=20 sin }( 30t-\frac{\pi }{4} ) $ . In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively: [JEE Main Online 08-04-2018]

#### Options:

A) $ \frac{50}{\sqrt{2}},0 $

B) $ 50,0 $

C) $ 50,10 $

D) $ \frac{1000}{\sqrt{2}},10 $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

$ e=100\sin (30t) $ $ i=20\sin ( 30t-\frac{\pi }{4} ) $

$ e _{rms}=\frac{100}{\sqrt{2}}volt, $

$ i _{rms}=\frac{20}{\sqrt{2}}amp $

Power factor $ =\cos \phi =cos( -\frac{\pi }{4} )=\frac{1}{\sqrt{2}} $

$ <P\gt=e _{rms}i _{rms}\cos \phi $

$ =\frac{100}{\sqrt{2}}\times \frac{20}{\sqrt{2}}\times \frac{1}{\sqrt{2}} $

$ =\frac{1000}{\sqrt{2}}watt $ Wattles current $ =i _{rms}\sin \phi $

$ =\frac{20}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=10amp $