The current flowing through a wire of resistance 2 Ω varies with time as shown in figure alongside. The amount of heat produced

(in J) in 3 would be

Assuming that each interval is of 1 seconds, we can calculate the heat for each interval and just sum it up to find the net heat produced,

Heat, H= I² RT

Where,

I is the current,

R is the resistance,

T is the time taken

For AD,

I=3, R=2 Ω, T=1 sec

Putting the values in the above formula, we get

H_AD=3² . 2 . 1=18 J

For DG,

I=-2 A, R= 2Ω, T=1 sec

Putting the values in the above formula, we get

H_DG=(-2)² . 2 . 1=8 J

For GJ,

I=1 A, R=2 Ω, T=1 sec

Putting the values in the above equation, we get

H_GJ=1² . 2 . 1=2 J

So Total amount of heat generated in 3 sec

H_NET = 18 + 8 + 2 = 28 J