# Power Distribution

The discussion of series circuits that the power applied to a series resistive circuit equals the power dissipated by the resistive elements. The same is true for parallel resistive networks. In fact,
For any network composed of resistive elements, the power applied by the source will equal that dissipated by the resistive elements.
For the parallel circuit in Fig. No.1: $$\bbox[5px,border:1px solid red] {\color{blue}{P_E = P_{R1} + P_{R2} + P_{R3}}} \tag{1}$$
Fig.No.1: Power flow in a dc parallel network.
The power delivered by the supply can be determined using $$\bbox[5px,border:1px solid red] {\color{blue}{P_E = EI_s \text{(watts, W)}}} \tag{2}$$ The power dissipated by the resistive elements can be determined by any of the following forms (shown for resistor R1 only): $$\bbox[5px,border:1px solid red] {\color{blue}{P_1 = V_1 \times I_1 = V_1 \times {V_1 \over R_1} = {{V_1}^2 \over R1} \text{(watts, W)}}} \tag{3}$$ In the equation $P = V_2/R$, the voltage across each resistor in a parallel circuit will be the same. The only factor that changes is the resistance in the denominator of the equation. The result is that
in a parallel resistive network, the larger the resistor, the less is the power absorbed.
Example 1: For the parallel network in Fig. No.2 (all standard values):
a. Determine the total resistance $R_T$.
b. Find the source current and the current through each resistor.
c. Calculate the power delivered by the source.
d. Determine the power absorbed by each parallel resistor.
e. Verify Eq. 1.
Fig.no.2: Parallel network for Example 1.
Solution:
a. Without making a single calculation, it should now be apparent from previous examples that the total resistance is less than 1.6 kΩ and very close to this value because of the magnitude of the other resistance levels: $$R_T = {1 \over {{1 \over R1}+{1\over R2}+{1\over R3}}}$$ $$={ 1\over {{1\over1.6 kΩ}+{1\over 20 kΩ} + {1\over 56 kΩ}}}$$ $$= {1 \over {625 \times 10^{-6} + 50 \times 10^{-6} + 17.867 \times 10^{-6}}}$$ $$= {1 \over {692.867 \times 10^{-6}}}$$ $$R_T = 1.44 kΩ$$
b. Applying Ohm's law gives $$Is = {E \over R_T}$$ $$Is = {28 V \over 1.44 kΩ}= 19.44 mA$$
Applying Ohm's law again gives $$I_1 ={ V_1 \over R_1} = {28 V \over 1.6 kΩ}= 17.5 mA$$ $$I_2 ={ V_2 \over R_2} = {28 V \over 20 kΩ}= 1.4 mA$$ $$I_3 ={ V_3 \over R_3} = {28 V \over 56 kΩ}= 0.5 mA$$
c. Applying Eq. (1) gives $$P_E = EIs = (28 V)(19.4 mA) = 543.2 mW$$ d. Applying each form of the power equation gives $$P_1 = V_1 I_1 = E I_1 = (28 V)(17.5 mA) = 490 mW$$ Similarly $$P_2 = 39.2 mW$$ and $$P_3 = 14 mW$$ A review of the results clearly substantiates the fact that the larger the resistor, the less is the power absorbed.
e. $$P_E = P_{R1} + P_{R2} + P_{R3}$$ $$543.2 mW = 490 mW + 39.2 mW + 14 mW = 543.2 mW$$