Electricity flashcards

3.5.1 Current electricity

Potential difference

Work done per unit charge (V = W/Q).

Emf vs pd

Emf: energy given to each coulomb by the source. Pd: energy transferred from each coulomb by a component.

Resistance on a curved I–V graph

R = V/I, valid at any point on any I–V graph. It is not the gradient.

Ohm's law

Current is proportional to pd under constant physical conditions (such as constant temperature). It is a special case, not the definition of resistance.

Explain the shape of a filament lamp's I–V graph

As current rises, the filament gets hotter, the metal ions vibrate more, and resistance rises, so the gradient of I against V falls. The graph is symmetrical for reversed pd.

A wire's length is doubled and its diameter doubled. What happens to its resistance?

R = ρL/A and A ∝ d², so R ∝ L/d². Doubling L doubles R; doubling d divides it by 4. Overall R halves.

ntc thermistor

Resistance decreases as temperature increases (a falling R–T curve). Used as a temperature sensor.

Superconductor

Zero resistivity at and below a critical temperature, which depends on the material. Uses: producing strong magnetic fields; less energy loss in power transmission.

Cells in series vs n identical cells in parallel

Series: emfs add (watch their direction), internal resistances add. n identical cells in parallel: emf of one cell, internal resistance r/n.

Potential divider with a thermistor: what happens to its pd as it warms?

Hotter → its resistance falls → its share of the pd falls. To make V_out rise with temperature, take it across the fixed resistor.

Required practical 6: finding emf and internal resistance from a graph

Vary an external resistor and record terminal pd V and current I. Plot V against I: y-intercept = ε, gradient = −r (from V = ε − Ir).

Terminal pd is 5.6 V at 0.40 A and 4.8 V at 1.20 A. Find r and ε.

r = (5.6 − 4.8) ÷ (1.20 − 0.40) = 1.0 Ω. ε = V + Ir = 5.6 + 0.40 × 1.0 = 6.0 V.