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. 1.
Fig. 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,
$$\bbox[10px,border:1px solid grey]{P_E = P_{R_1} + P_{R_2} + P_{R_3}} \tag{1}$$
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):
$$P_1 = V_1 \times I_1 = I_1R_1 \times I_1 $$
$$\bbox[10px,border:1px solid grey]{P_1 = V_1 \times I_1 = I_1^2R_1} \, \text{(watts, W)} \tag{3}$$
$$\bbox[10px,border:1px solid grey]{P_1 = V_1 (V_1 / R_1) = {V_1^2 \over R_1}} \, \text{(watts, W)} \tag{4}$$
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. 2:
a. Calculate the resulting source current $I$.
b. Determine the power dissipation in each resistor.
Fig. 2: Series circuit to be analyzed in the example 1.
a: $E = I \times R_T$
so $$R_T = R_1 + R_2 +R_3$$ $$R_T = 2+3+5 = 10Ω$$ Now $$I = {E \over R_T} ={10 \over 10} = 1A$$ b: $$P_T = P_1 + P_2 +P_3$$ $$ P_1 = I^2R_1 = (1)^2(2) = 2 \text{ watts} $$ $$ P_2 = I^2R_2 = (1)^2(3) = 3 \text{ watts}$$ $$ P_3 = I^2R_3 = (1)^2(5) = 5 \text{ watts}$$ So $$ P_T= 2 + 3 +5 = 10 \text{ watts}$$

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