3.4 Power dissipation in resistor circuits

We can find voltages and currents in simple circuits containing resistors and voltage or current sources.We should examine whether these circuits variables obey the Conservation of Power principle:since a circuit is a closed system, it should not dissipate or create energy.For the moment, our approach is to investigate first a resistor circuit's power consumption/creation. Later, we will prove that because of KVL and KCL all circuits conserve power.

As defined on [link] , the instantaneous power consumed/created by every circuit element equals the product of itsvoltage and current. The total power consumed/created by a circuit equals the sum of eachelement's power. $$P=\sum_k v_k{i}_k$$ Recall that each element's current and voltage must obey the convention that positive current is defined to enter the positive-voltage terminal.With this convention, a positive value of $v_k{i}_k$ corresponds to consumed power, a negative value to created power. Because the total power in a circuit must be zero( $P=0$ ), some circuit elements must create power while others consume it.

Consider the simple series circuit should in [link] . In performing our calculations, we defined the current $i_{\text{out}}$ to flow through the positive-voltage terminals of both resistors and found it to equal $i_{\text{out}}=\frac{v_{\text{in}}}{R_1+R_2}$ . The voltage across the resistor $R_2$ is the output voltage and we found it to equal $v_{\text{out}}=\frac{R_2}{R_1+R_2}{v}_{\text{in}}$ . Consequently, calculating the power for this resistor yields $$P_2=\frac{R_2}{(R_1+R_2)^2}{v}_{\text{in}}^2$$ Consequently, this resistor dissipates power because $P_2$ is positive. This result should not be surprising since we showed that the power consumedby any resistor equals either of the following.

We conclude that both resistors in our example circuit consume power, which points to the voltage source as the producer of power.The current flowing into the source's positive terminal is $-i_{\text{out}}$ . Consequently, the power calculation for the source yields $$-(v_{\text{in}}{i}_{\text{out}})=-(\frac{1}{R_1+R_2}{v}_{\text{in}}^2)$$ We conclude that the source provides the power consumed by the resistors, no more, no less.

This result is quite general: sources produce power and the circuit elements, especially resistors,consume it. But where do sources get their power?Again, circuit theory does not model how sources are constructed, but the theory decrees that all sources must be provided energy to work.