The power dissipated (in watt ) in 3 Ω resistance in the following circuit is

Step-by-step solution

From circuit, 1 R = 1 3 + 1 6 = 2 + 1 6 = 3 6 Ω ⇒ R 1 = 2 Ω R 2 = ( R 2 + R 2 ) ∥ R 4 = 2 Ω The internal resistance of battery = 1 Ω So, the equivalent resistance of circuit =2 Ω + 1 Ω =3 Ω The current in the battery, I = V R = 4 .5 3 =1.5 A Current in 3 Ω is, 1 .5 2 × 2 3 A = 1 2 A The power dissipated in 3 Ω resistance, P = I 2 × R = 1 2 2 × 3 = 1 4 × 3 = 0 .75 W