Question

what is the approximate peak value of an alternating current producing 4 times the heat produced per second by a steady current of 2A in a resistor

1) 2.8A

2) 4A

3) 5.6A

4) 8A

Answer 1

Irms= 4×i steady (1)
Also,irms =Imax/√2. (2)
From 1 &2
Imax/√2=4×i steady
So Imax =5.6 A

Answer 2

Dear Student,

◆ Answer -

(3) 5.6 A

● Explanation -

Given that heat produced by AC is same as DC.

Heat (AC) = 4 × Heat (DC)

Irms^2.R.t = 4 × I^2.R.t

Irms^2 = 4I^2

Irms = 2I

Irms = 2 × 2

Irms = 4 A

Peak value of AC is given as -

Ipeak = √2 Irms

Ipeak = 1.414 × 4

Ipeak = 5.656 A

Hence, answer is (3) 5.6 A.

Thank dear. Hope this helps you..