- 28J
Explanation:
Assuming that each interval is of 1seconds, we can calculate the heat for each interval and just sum it up to find the net heat produced,
Heat, H = I2RT
Where,
I is the current,
R is the resistance,
T is the time taken
For AD,
$\text{I}=3, \text{R}=2\Omega,\text{T}=1\text{sec}$
Putting the values in the above formula, we get
$\text{H}_{\text{AD}}=3^{2}.2.1=18\text{J}$
For DG,
$\text{I}=-2\text{A},\text{R}=2\Omega,\text{T}=1\text{sec}$
Putting the values in the above formula, we get
$\text{H}_{\text{DG}}=3^{2}.2.1=18\text{J}$
For GJ,
$\text{I}=-2\text{A},\text{R}=2\Omega,\text{T}=1\text{sec}$
Putting the values in the above equation, we get
$\text{H}_{\text{GJ}}=1^{2}.2.1=2\text{J}$
So Total amount of heat generated in 3sec
$\text{H}_{\text{NET}}=18+8+2=28\text{J}$