In a electric circuit where two resistances of 1 ohm and 3 ohm connected in series and 8 ohm connected in parallel The power dissipated across 8 ohm is 2 watt What is the power dissipated in 3 ohm?

Answer 1

First thing you need to do is calculate the Voltage across the 8 ohm resistor. As it is in parallel with the 1 & 3 ohm resistors in series that would be the voltage across those two. P = E^2 / R thus P x R = E^2; 2 watt x 8 ohms = E^2 = 4V Quick check: I = E / R = 4 / 8 = 0.5 A P = E x I = 4 x 0.5 = 2 watts Now Rt = 1 ohm + 3 ohms = 4 ohms Caclulate current in the series circuit: I = E / R = 4V / 4 ohms = 1A Now calaculate the power in the 3 ohm resistor: P = I^2 x R = 1 x 3 = 3 Watts. (NOTE: I am using the ^2 to show exponents, in this case it means squared)