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Encyclopedia of Electrical Engineering

# 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,
 $$\bbox[5px,border:1px solid red] {\color{blue}{P_E = P_{R1} + P_{R2} + P_{R3}}}$$ $$\text{(eq.1)}$$
The power delivered by the supply can be determined using
 $$\bbox[5px,border:1px solid red] {\color{blue}{P_E = EI_s \text{(watts, W)}}}$$ $$\text{(eq.2)}$$
The power dissipated by the resistive elements can be determined by any of the following forms (shown for resistor R1 only): $$P1 = V1 \times I1 = I1R1 \times I1$$
 $$\bbox[5px,border:1px solid red] {\color{blue}{P1 = V1 \times I1 = I1^2R1 \text{(watts, W)}}}$$ $$\text{(eq.3)}$$
or
 $$\bbox[5px,border:1px solid red] {\color{blue}{P1 = V1 (V1 / R1) = {V1^2 \over R1} \text{(watts, W)}}}$$ $$\text{(eq.4)}$$
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
$P_T= 2 + 3 +5 = 10 watts$