Hint : Try to find current flowing in the
4Ω resistor then use the formula for power dissipated in a resistor.
Step 1 : Find equivalent resistance of the circuit.
Given: Two resistors of 8Ω in parallel.
One resistance of 4Ω in series with above combination.
Net resistance of two two resistors of 8Ω
1Rnet= 1R1+1R2
1Rnet= 18+18
⟹ Rnet=4Ω
Equivalent resistance of whole circuit
Req=R4Ω+Rnet
⟹ Req=4Ω+4Ω=8Ω
Step 2 : Find current flowing in the circuit using Ohm's Law.
From Ohm's Law, V=IReq
I=8V8Ω=1 A.
This current will flow through the 4Ω resistor.
Step 3 : Find power dissipated in 4Ω resistor.
Power dissipated across 4Ω resistor is
P=I2R4Ω
P=(1A)2×4Ω=4W
Final Step : The Power dissipated across the 4Ω resistor is 4W.