An alternating voltage v(t)=220\; \sin 100\; \pi t volt is applied to a purely resistive load of 50\Omega. The time taken for the current to rise from half to the peak value to the peak value is :

  • Option 1)

    7.2\; ms

  • Option 2)

    2.2\; ms

  • Option 3)

    5\; ms

     

  • Option 4)

    3.3\; ms

Answers (1)

V=220\; \sin (100\: \pi\: t)

w=100\: \pi

T=\frac{2\pi }{w}=\frac{2\pi }{100\pi }=\frac{1}{50}sec.

R=50\Omega

Let time taken from half of peak value to the peak value =\Delta t

\Delta t=\frac{T}{6}=\frac{1}{50\times 6}=\frac{1}{300}sec

\Delta t=\frac{10}{3}ms=3.3 \; ms
 


Option 1)

7.2\; ms

Option 2)

2.2\; ms

Option 3)

5\; ms

 

Option 4)

3.3\; ms

Preparation Products

Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions