# Series Circuit Power Distribution

In any electrical system, the power applied will equal the power dissipated or absorbed.
For any series circuit, such as that in Fig. no.1. Fig.no.1: Power distribution in series circuit.
The power applied by the dc supply must be equal to the power dissipated by the resistive elements in the circuit.
In equation form, $$\bbox[10px,border:1px solid grey]{P_E = P_{R1} + P_{R2} + P_{R3}} \tag{1}$$ The power delivered by the supply can be determined using $$\bbox[10px,border:1px solid grey]{P_E = EI_s} \, \text{(watts, W)} \tag{2}$$ The power dissipated by the resistive elements can be determined by any of the following forms (shown for resistor R1 only): $$P_1 = V_1 \times I_1 = I_1R_1 \times I_1$$ $$\bbox[10px,border:1px solid grey]{P_1 = V_1 \times I_1 = I_1^2R_1} \, \text{(watts, W)} \tag{3}$$ or $$\bbox[10px,border:1px solid grey]{P_1 = V_1 (V_1 / R_1) = {V_1^2 \over R_1}} \, \text{(watts, W)} \tag{4}$$ Since the current is the same through series elements, you will find in the following examples that
in a series configuration, maximum power is delivered to the largest resistor.
Example 1:
For the series circuit in Fig. no.2:
a. Calculate the resulting source current $I$.
b. Determine the power dissipation in each resistor. Fig.no.2: Series circuit to be analyzed in the example 1.
Solution:
a:
$E = I \times R_T$
so
$R_T = R1 + R2 +R3$
$R_T = 2+3+5 = 10Ω$
$R_T = 10Ω$
Now
$I = {E \over R_T} ={10 \over 10} = 1A$
b:
$P_T = P_1 + P_2 +P_3$
$P1 = I^2R1 = (1)^2(2) = 2 watts$
$P2 = I^2R2 = (1)^2(3) = 3 watts$
$P3 = I^2R3 = (1)^2(5) = 5 watts$
So
$P_T= 2 + 3 +5 = 10 watts$