Guide 21-4. Power and Energy in Series and Parallel Circuits

In Guide 21-2, you saw how to apply the conservation laws to determine equivalent resistance formulas for series and parallel circuits. In this guide, you'll see how to apply the laws to power and energy in series and parallel circuits. We'll look at the energy produced in the battery and then converted to thermal energy in the resistors. (There would also be radiant energy if the resistor were a bulb.)

 Power and Energy in a Series Circuit

Considering once again the series circuit of 2 resistors (shown to the right for convenience),  ΔU_b + ΔU₁ + ΔU₂ = 0 or ΔU_b = -(ΔU₁ + ΔU₂). The minus sign is needed, because energy is gained by charge in the battery and lost in the resistors.  In a given time interval, Δt,

ΔU_b/Δt = -(ΔU₁ + ΔU₂)/Δt.

Substituting potential differences, we have QΔV_b/Δt = -(QΔV₁ + QΔV₂)/Δt.

Using the definition of current, the equation becomes IΔV_b = -IΔV₁ - IΔV₂. We use the same I in all terms, because the current is the same in all parts of a series circuit. 

Substituting ΔV₁ = -IR₁ and ΔV₂ = -IR₂ gives IΔV_b = I²R₁ + I²R₂.

Equivalently, we could have substituted I = -ΔV₁/R₁ = -ΔV₂/R₂ to obtain IΔV_b = (-ΔV₁)²/R₁ + (-ΔV₂)²/R₂ = (ΔV₁)²/R₁ + (ΔV₂)²/R₂.

Now let's examine three of the previous relationships together:

ΔU_b/Δt = -(ΔU₁ + ΔU₂)/Δt

IΔV_b = I²R₁ + I²R₂

IΔV_b = (ΔV₁)²/R₁ + (ΔV₂)²/R₂

Recognizing that the rate of change of energy is just power, P, all of these relationships are equivalent to the following relationship between the rate of energy production in the battery and the rates of energy dissipation in the resistors:

P_b = P₁ + P₂.

In writing the above formula, we use the textbook convention that the symbol P represents a positive value. Therefore, P_b, represents a power gain in the battery, while P₁ and P₂ represent power losses in the resistors.

 Power and Energy in a Parallel Circuit

Now consider energy in the parallel circuit shown to the right.  Starting with the junction rule: I = I₁ + I₂.

Let's rewrite this as:  I = -ΔV₁/R₁ - ΔV₂/R₂.  Now P = (ΔV_r)²/R, so P/ΔV_r= ΔV_r/R. Making that substitution for the terms on the left side of the equation: I = -P₁/ΔV₁ - P₂/ΔV₂.

Since this is a parallel circuit, we know that ΔV_b = -ΔV₁ = -ΔV₂.  Substituting into the previous equation, we have I = P₁/ΔV_b + P₂/ΔV_b.

Multiplying through by ΔV_b, IΔV_b = P₁ + P₂.

Note that IΔV_b = QΔV_b/Δt = ΔU_b/Δt = P_b. Therefore, we obtain once again P_b = P₁ + P₂. The fact that this is the same result as for the series circuit makes sense from the standpoint of energy conservation. All the energy that's produced in the battery is converted to other forms in the resistors.  That's independent of how the resistors are wired into the circuit. However, the way that the energy is distributed between the resistors does depend on whether they're in series or parallel. This will be investigated later in a video exercise on series and parallel light bulbs.

 Summary

What we've found so far is one simple formula that applies to both series and parallel circuits of two resistors and a battery. We could extend the argument to any number of resistors in series and parallel circuits and combinations of such circuits.

A consequence of the law of conservation of energy applied to circuits is that all of the electrical energy provided by the battery is converted to other forms by the circuit components. For a circuit of two resistors and a battery, we express this in terms of power as Pb = P1 + P2.

 Comparing Power Dissipation in Series and Parallel Circuits

The current provided by an energy source increases or decreases depending on the equivalent resistance of the circuit. Consider the following three cases for two resistors of identical resistance, R. The potential difference across the battery is ΔV_b in every case. However, the total currents, I, will be different, because the equivalent resistance of the circuit is different for each case.

Case 1	Case 2	Case 3

Case 1.  The equivalent resistance is simply R_eq = R.  Thus, I = ΔV_b/R_eq = ΔV_b/R. The power dissipated by the resistor is P_r = I²R_eq = (ΔV_b)²/R.

Case 2.  The equivalent resistance is R_eq = 2R.  Thus, I = ΔV_b/R_eq = ΔV_b/(2R) = ½(ΔV_b/R). The total current is half as much as for Case 1. The total power dissipated by both resistors together is P_r= I²R_eq = [½(ΔV_b/R)]²(2R) = ½(ΔV_b)²/R. So that's half as much as for Case 1. Note that each resistor alone dissipates a quarter as much power as the resistor in Case 1.

Case 3.  The equivalent resistance is R_eq = R/2.  Thus, I = ΔV_b/R_eq = ΔV_b/(R/2) = 2(ΔV_b/R). The total current is twice as much as for Case 1. The total power dissipated by both resistors together is P_r= I²R_eq = [2(ΔV_b/R)]²(R/2) = 2(ΔV_b)²/R. So that's twice as much as for Case 1. Note that each resistor alone dissipates the same power as the resistor in Case 1.

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