The correct option is A 4 W.
The circuit can be reduced, step by step, to a single equivalent resistance. The 8 ohm and the 8 ohm are connected in parallel , and so they can be replaced by an equivalent resistor Rp of 4 ohms, using the below equation.
Rp=(8×8)(8+8)=4ohms
This resistor is connected in series with the 4 ohm resistor R1. The total resistance Rt of the circuit is then
Rt =R1+Rp=4+4 = 8 ohms.
Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get,
I=VR=88=1A.
Hence, the current flowing through the resistor R1 (4 ohms) is 1A.
Now the power dissipated in the resistor R1 (4 ohms) is given as follows.
P=I2R=12×4=1×4=4W.
Hence, the power dissipated in the resistor R1 (4 ohms) = 4W.