

Guide 214. Power and Energy in Series and Parallel Circuits In Guide 212, 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} + ΔU_{1} + ΔU_{2} = 0 or ΔU_{b} = (ΔU_{1} + ΔU_{2}). 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 = (ΔU_{1} + ΔU_{2})/Δt. Substituting potential differences, we have QΔV_{b}/Δt = (QΔV_{1} + QΔV_{2})/Δt. Using the definition of current, the equation becomes IΔV_{b} = IΔV_{1}  IΔV_{2}. We use the same I in all terms, because the current is the same in all parts of a series circuit. Substituting ΔV_{1} = IR_{1} and ΔV_{2} = IR_{2} gives IΔV_{b} = I²R_{1} + I²R_{2}. Equivalently, we could have substituted I = ΔV_{1}/R_{1} = ΔV_{2}/R_{2} to obtain IΔV_{b} = (ΔV_{1})²/R_{1} + (ΔV_{2})²/R_{2} = (ΔV_{1})²/R_{1} + (ΔV_{2})²/R_{2}. Now let's examine three of the previous relationships together: ΔU_{b}/Δt = (ΔU_{1} + ΔU_{2})/Δt IΔV_{b} = I²R_{1} + I²R_{2} IΔV_{b} = (ΔV_{1})²/R_{1} + (ΔV_{2})²/R_{2} 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} = P_{1} + P_{2}. 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 P_{1} and P_{2} 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 = I_{1} + I_{2}. Let's rewrite this as: I = ΔV_{1}/R_{1}  ΔV_{2}/R_{2}. 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 = P_{1}/ΔV_{1}  P_{2}/ΔV_{2}. Since this is a parallel circuit, we know that ΔV_{b} = ΔV_{1} = ΔV_{2}. Substituting into the previous equation, we have I = P_{1}/ΔV_{b} + P_{2}/ΔV_{b}. Multiplying through by ΔV_{b}, IΔV_{b} = P_{1} + P_{2}. Note that IΔV_{b} = QΔV_{b}/Δt = ΔU_{b}/Δt = P_{b}. Therefore, we obtain once again P_{b} = P_{1} + P_{2}. 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.
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. 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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