Question:

An alternating voltage V(t) = 220 sin 100 Ωt 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:

(1) 5 ms 

(2) 3.3 ms

(3) 7.2 ms 

(4) 2.2 ms

                                        JEE-2024

Solution:                
An alternating voltage V(t) = 220 sin 100 Ωt 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: (1) 5 ms (2) 3.3 ms (3) 7.2 ms (4) 2.2 ms JEE-2024

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