Q80 / 90 · JEE Main 2024 (30 Jan Shift 2) · Physics

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 the peak value to the peak value is:

A
$5 ms$
Correct answer
B
$3.3 ms$
C
$7.2 ms$
D
$2.2 ms$

Solution

Time taken to change current value from half of the peak value to the peak value is,
$t = \frac{T}{6}$
$\Rightarrow t = \frac{2 π}{6 ω} = \frac{π}{3 ω} = \frac{π}{300 π} = \frac{1}{300} = 3.33 ms$

Hence, the correct answer is option: 'B'.

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An alternating voltage V(t)=220sin100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:

Solution

Relevant Questions

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 the peak value to the peak value is: