Electrical power and I-V characteristicsEdexcel International A Level Physics: Revision notes

Power is the rate of energy transfer, $P = W/t$. Since $V = W/Q$ and $I = Q/t$, the power transferred in a component is

$P = VI$

and the energy transferred in time $t$ is $W = VIt$. Substituting $V = IR$ and $I = V/R$ (from the definition of resistance) gives two more forms:

$P = I^2R \qquad P = \frac{V^2}{R}$

Use $P = I^2R$ when the current is known (or the same through components in series) and $P = V^2/R$ when the p.d. is known (or the same across components in parallel).

An I–V characteristic plots current against potential difference for a component. For an ohmic conductor at constant temperature (a fixed resistor or metal wire) it is a straight line through the origin, and the resistance $V/I$ is constant. A steeper line means a smaller resistance. The graph is symmetrical when the p.d. is reversed.

For a filament lamp the I–V graph is a curve through the origin whose gradient decreases as the p.d. increases. The current heats the filament, the metal ions vibrate with greater amplitude, and the conduction electrons collide with them more often, so the resistance rises with temperature. It is non-ohmic. The cold filament has a low resistance, so the current at switch-on is much larger than in normal operation.

An NTC thermistor (negative temperature coefficient) is made of semiconductor material. As its temperature rises, more charge carriers are released, so the number of carriers per unit volume increases and the resistance falls. The I–V graph is a curve through the origin whose gradient increases as the p.d. increases (the current heats the thermistor). At constant p.d., a falling resistance gives a larger power, so there is a risk of thermal runaway.

A semiconductor diode conducts in one direction only. In the forward direction the current is almost zero below a threshold p.d. of about 0.6 to 0.7 V, then rises very rapidly, so the resistance falls sharply. In the reverse direction the current is almost zero (very large resistance) until breakdown. The I–V graph is therefore not symmetrical. Because the forward current can be large, a protective resistor is often connected in series.

A 12 V, 36 W lamp has cold resistance 0.40 Ω.

• Operating current: $I = P/V = 3.0$ A
• Operating resistance: $R = V/I = 4.0$ Ω
• Energy in 2.0 min: $W = Pt = 36 \times 120 = 4.3 \times 10^3$ J
• Switch-on current: $12/0.40 = 30$ A, ten times larger