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), ΔUb + ΔU1 + ΔU2 = 0 or ΔUb = -(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,
ΔUb/Δt = -(ΔU1 + ΔU2)/Δt.
Substituting potential differences, we have QΔVb/Δt = -(QΔV1 + QΔV2)/Δt.
Using the definition of current, the equation becomes IΔVb = -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 IΔVb = I²R1 + I²R2.
Equivalently, we could have substituted I = -ΔV1/R1 = -ΔV2/R2 to obtain IΔVb = (-ΔV1)²/R1 + (-ΔV2)²/R2 = (ΔV1)²/R1 + (ΔV2)²/R2.
Now let's examine three of the previous relationships together:
ΔUb/Δt = -(ΔU1 + ΔU2)/Δt
IΔVb = I²R1 + I²R2
IΔVb = (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:
Pb = P1 + P2.
In writing the above formula, we use the textbook convention that the symbol P represents a positive value. Therefore, Pb, 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 = (ΔVr)²/R, so P/ΔVr= ΔVr/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 ΔVb = -ΔV1 = -ΔV2. Substituting into the previous equation, we have I = P1/ΔVb + P2/ΔVb.
Multiplying through by ΔVb, IΔVb = P1 + P2.
Note that IΔVb = QΔVb/Δt = ΔUb/Δt = Pb. Therefore, we obtain once again Pb = 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.
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.
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 ΔVb 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. The equivalent resistance is simply Req = R. Thus, I = ΔVb/Req = ΔVb/R. The power dissipated by the resistor is Pr = I²Req = (ΔVb)²/R.
Case 2. The equivalent resistance is Req = 2R. Thus, I = ΔVb/Req = ΔVb/(2R) = ½(ΔVb/R). The total current is half as much as for Case 1. The total power dissipated by both resistors together is Pr= I²Req = [½(ΔVb/R)]²(2R) = ½(ΔVb)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 Req = R/2. Thus, I = ΔVb/Req = ΔVb/(R/2) = 2(ΔVb/R). The total current is twice as much as for Case 1. The total power dissipated by both resistors together is Pr= I²Req = [2(ΔVb/R)]²(R/2) = 2(ΔVb)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.
© North Carolina School of Science and Mathematics, All Rights Reserved. These materials may not be reproduced without permission of NCSSM.