realnfo.com
Encyclopedia of Electrical Engineering
HOME TABLE OF CONTENTS PROJECTS TECHNOLOGY
HOME TABLE OF CONTENTS
realnfo.com
Encyclopedia of Electrical Engineering
HOME TABLE OF CONTENTS PROJECTS TECHNOLOGY
HOME TABLE OF CONTENTS

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$