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


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

  1. $5 \mathrm{~ms}$

  2. $2.2 \mathrm{~ms}$

  3. $7.2 \mathrm{~ms}$

  4. $3.3 \mathrm{~ms}$

Correct Option: 4

JEE Main Previous Year 1 Question of JEE Main from Physics Alternating Current chapter.
JEE Main Previous Year 8 April 2019 I


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