Define

(i) electromotive force (emf ) of a battery.

(ii) electrical resistance of a conductor.

A battery of emf ε and negligible internal resistance is connected in series to two resistors. The current in the circuit is I.

(i) State an equation giving the total power delivered by the battery.

(ii) The potential difference across resistor R₁ is V₁ and that across resistor R₂ is V₂.  Using the law of the conservation of energy, deduce the equation below.

ε =V₁ +V₂

The graph shows the I-V characteristics of two conductors, X and Y.

On the axes below, sketch graphs to show the variation with potential difference V of the resistance of conductor X (label this graph X) and conductor Y (label this graph Y). You do not need to put any numbers on the vertical axis.

The conductors in (c) are connected in series to a battery of emf ε and negligible internal resistance.

The power dissipated in each of the two resistors is the same.

Using the graph given in (c),

(i) determine the emf of the battery.

(ii) calculate the total power dissipated in the circuit.

(i) the work done per unit charge in moving a quantity of charge completely around a circuit / the power delivered per unit current / work done per unit charge made available by a source; 

(ii) the ratio of the voltage (across) to the current in the conductor;

(i) emf × current;

(ii) total power is V₁I +V₂I;
equating with EI to get result;
or
total energy delivered by battery is EQ;
equate with energy in each resistor V₁Q +V₂Q;

graph X: horizontal straight line;
graph Y: starts lower than graph X;
rises (as straight line or curve) and intersects at 4.0 V;

Do not pay attention to numbers on the vertical axis.

(i) realization that the voltage must be 4.0 V across each resistor;
and so emf is 8.0 V;

(ii) power in each resistor = 3.2W;
and so total power is 6.4 W;
or
current is 0.80 A;
so total power is 8.0×0.80 = 6.4W;