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# Illustration 31.5: Power and Reactance

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Assume an ideal power supply. The graph shows the voltage **(red)** across the source and the current **(black)** through the circuit as a function of time **(voltage is given in volts, current is given in milliamperes, and time is given in seconds)**. Restart.

Resistive circuit: Look at the plot of voltage and current. Power is given by P = VI, but the current and voltage vary in time. It is more useful to think about the average power, which is P = V_{rms}I_{rms} = I_{rms}^{2}R = V_{rms}^{2}/R. Notice that the current and voltage are always in phase, and so the product VI is always positive.

Capacitive circuit: Look at the plot of voltage and current. Notice that when the voltage is going from 0 to a more positive number, the current is going from a maximum value toward 0, but then when the voltage is going from its max back towards 0, the current has changed direction and is going from 0 down to a negative value. Thus, on the average over many cycles, the current and voltage are out of phase by π/2 = 90^{o}. When the voltage is positive, the current is negative as much as it is positive, and the same applies when the voltage is negative. This means that the average power is 0. Compare this with the resistive load. When the voltage is positive, the current is positive, and when the voltage is negative, the current is negative. The resistor is always drawing current away from the source. Another way to think about this is that the capacitor simply stores charge. As the voltage changes direction, the current goes back and forth between the source and the capacitor, so the capacitor does not dissipate any energy over time (it simply stores the energy briefly).

Inductive circuit: Compare the plot of voltage and current for the inductor to that of the capacitor. Can you explain why the average power is 0 for this inductor just as it is for the capacitor?

If you have a circuit with a combination of resistive, capacitive, and inductive loads, calculating the average power dissipated now requires calculating V_{rms}I_{rms}cosφ, where φ is the phase shift between the current and the voltage (see Exploration 31.4).

Illustration authored by Anne J. Cox.

Script authored by Wolfgang Christian and Anne J. Cox.

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