UY1: Electromotive Force & Power In Circuits

In this post, we will do a short recap on emf and power in electric circuits.


Consider a circuit element with potential difference $V_{a} – V_{b} = V_{ab}$ between its terminals (a and b) and current $I$ passing through it in the direction from a towards b.


As charge passes through the circuit element, the electric field does work on the charge. The moving charges do not gain kinetic energy. The rate at which energy is delivered to (or extracted from) a circuit element is the power:

$$P = IV_{ab}$$

If the circuit element is a resistor, then:

$$V_{ab} = IR$$

This means the power is:

$$P = I^{2}R = \frac{V_{ab}^{2}}{R}$$


For a conductor to have a steady current, it must be part of a conducting path that forms a closed loop or complete circuit. There must be a source of electromotive force (emf) to maintain a steady current in a complete circuit.

An ideal source of emf maintains a constant potential difference $V_{ab}$ between its terminals, independent of the current through it. We define electromotive force $E$ quantitively as the magnitude of this potential difference:

$$E = V_{ab}$$

Note that for a real source of emf, (due to internal resistance)

$$V_{ab} < E$$

Assuming Ohmic behaviour, (after accounting for internal resistance, r)

$$E = V_{ab} + Ir$$

$$I = \frac{V_{ab}}{R} = \frac{E}{R + r}$$


Power output of a source is the rate at which energy is extracted from the source of emf,

$$P = IV_{ab}$$

This is the rate of energy delivered to the external circuit. (Substituting $E = V_{ab} + Ir$)

$$P = EI – I^{2}r$$


Next: RC Circuits

Previous: Resistance of A Cylindrical Resistor

Back To Electromagnetism (UY1)

Back To University Year 1 Physics Notes

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