Physics · 2019
JEE Main · 8 April 2019, Shift 1 · Q19
An alternating voltage v( t )=220 sin 100 π t volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of…
An alternating voltage $\displaystyle v(\mathrm{t})=220 \sin 100 \pi \mathrm{t}$ volt is applied to a purely resistive load of $\displaystyle 50 \boldsymbol{\Omega}$. The time taken for the current to rise from half of the peak value to the peak value is :
Official answer
(2)
3.$\displaystyle 3$ ms
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.