When the switch is off, the current through the 4 Ω resistor is I = 12 V/12 Ω = 1 A
Power dissipated = I2R = 4 W
Now, when the switch is closed, let the current in various branches of the circuit be as shown in the image.
Applying KVL in loop 1
\(4I_1+2I_2 = 4\)
\(I_2 = 2 - 2I_1\ ...(i)
\)
Applying KVL In loop 2
\(-2I_2 + 8I_1 - 8I_2 = 0\)
\(-5I_2+4I_1 = 4\ .... (ii)\)
from (i) and (ii)
\(10 - 10I_1 + 4I_1 = 4\)
This gives,
\(I_1 = 1 \ A\)
The current through the 4 Ω resistor is the same even after the switch is closed, so there will be no change in power dissipated in the 4 Ω resistor.
