Power Distribution In Series Circuits:

Whenever computing the power in a circuit having resistors in series, all you require to do is find out the current I, in amperes, which the circuit is carrying. Then it is easy to compute the power P_n, in watts, dissipated by any specific resistor of value R_n, in ohms, by using the formula P_n = I²R_n.

The whole power dissipated in a series circuit is equivalent to the sum of the wattages dissipated in each resistor. In this manner, the distribution of power in a series circuit is like the distribution of the voltage.

PROBLEM:

Assume that we have a series circuit with a supply of 150 V and three resistors: R₁ = 330 ohms, R₂ = 680 ohms and R₃ = 910 ohms. Determine the power dissipated by R₂?

SOLUTION:

At first, we have to compute the total resistance. Since the resistors are in series, therefore total resistance, R = 330 + 680 + 910 = 1920 ohms. And hence, the current I = 150/1920 = 0.07813 A = 78.1 mA. The power dissipated by R₂ is

P₂ = I²R₂ = 0.07813 x 0.07813 x 680 = 4.151 W

We should round this off to three significant figures, obtaining 4.15 W.