## Q. 1.P.17

Find the power dissipation in 3 Ω resistor.

## Verified Solution

Applying KCL at mesh 2, we get

$-2 i_2-3\left(i_2-i_3\right)-1\left(i_2-i_1\right)=0$        (i)

The combined mesh equations for meshes (1) and (3) are

$\text { (7) }\left(i_1-i_2\right)-3\left(i_3-i_2\right)-1 . i_3=0$       (ii)

Solving, we get $i_3=2 A , i_1=9 A , i_2=2.5 A$

Inside the super mesh, applying KCL, we get

$i_1=7+i_3-7-i_3+i_1=0$

$i_1-i_3=7$            (iii)

So, the power dissipation is

$P=\left(i_2-i_3\right)^2 \times 3=(0.5)^2 \cdot 3 \Rightarrow P=0.75 W$

Note: Since the voltage across an ideal current source can be of any value, it is not possible to write the mesh equations for the meshes (1) and (2) independently.
Hence, the supemesh procedure is followed.