Circuits, EMF and power

MDCAT Physics · Physics · Current electricity. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

The EMF of a source is the energy it gives to each coulomb of charge, changing other forms of energy into electrical energy. The terminal potential difference is what is left for the outside circuit. It is smaller than the EMF when a current flows, because some energy is lost in the internal resistance of the source. On open circuit (no current) the two are equal.

Kirchhoff's first rule is the conservation of charge at a junction. His second rule is the conservation of energy round a closed loop.

A Wheatstone bridge compares resistances. At balance, the two ends of the galvanometer are at the same potential, so it carries no current.

Rules to remember

Terminal p.d. = EMF − lost volts: V = E − Ir
Current from a cell: I = E/(R + r)
power = p.d. × current: P = VI = I²R = V²/R
energy = power × time; 1 kW h = 3.6 × 10⁶ J
Kirchhoff's first rule: total current into a junction = total current out
Kirchhoff's second rule: round any closed loop, sum of EMFs = sum of IR drops; EMFs that oppose are subtracted
Balanced Wheatstone bridge: P/Q = R/S

Common mistake

Using the EMF as the potential difference across the external resistor. When a current flows, subtract the lost volts Ir first.

Worked example

A cell of EMF 6.0 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Find the terminal potential difference and the power in the resistor.

I = E ÷ (R + r) = 6.0 ÷ (2.5 + 0.50) = 2.0 A.
Terminal p.d. V = E − Ir = 6.0 − 2.0 × 0.50 = 5.0 V.
Power in the resistor P = VI = 5.0 × 2.0 = 10 W (check: I²R = 4.0 × 2.5 = 10 W).

Answer: 5.0 V and 10 W

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