(a) (i) Explain electromotive force (ii) list two sources of electromotive force other than a chemical cell (b) A chemical cell of electromotive force, E, a...

(a)(i) Electromotive force (e.m.f.): the total energy supplied by a source in driving one coulomb of charge completely round the whole circuit (both the external load and the internal resistance). Equivalently, it is the terminal potential difference of the source when no current is being drawn from it (open circuit). It is measured in volts (V).

(a)(ii) Two sources of e.m.f. other than a chemical cell:

• a dynamo / electrical generator (electromagnetic induction);
• a thermocouple (thermoelectric effect).

(A solar cell / photocell is also acceptable.)

(b) Circuit diagram. The cell (e.m.f. \(E\), internal resistance \(r\)) is joined in series with a plug key, an ammeter and the external load of resistance \(R\), while the voltmeter is connected across the terminals of the cell:

(c)(i) Physical meaning of \(V_o\) and \(V\):

• With the key open, no current flows, so there is no voltage drop across the internal resistance \(r\). The voltmeter reading \(V_o\) is therefore the full e.m.f. of the cell, \(V_o = E\).
• With the key closed, a current \(I\) flows and a voltage \(Ir\) is lost inside the cell. The voltmeter reading \(V\) is then the terminal potential difference of the cell, \(V = E - Ir\), which is less than \(V_o\).

(c)(ii)(I) Current in the circuit. Applying the circuit equation \(E = I(R+r)\):

\[ I = \frac{E}{R + r} \]

(c)(ii)(II) Maximum power dissipated. The power delivered to the external load is

\[ P_R = I^2 R = \frac{E^2 R}{(R + r)^2}. \]

Differentiating \(P_R\) with respect to \(R\) and setting \(\dfrac{dP_R}{dR}=0\) shows that this is greatest when the load is matched to the cell, i.e. when \(R = r\) (the maximum-power-transfer condition). Substituting \(R = r\):

\[ P_{R(\max)} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}. \]

At this same condition the power dissipated inside the cell is \(P_r = I^2 r = \dfrac{E^2 r}{(2r)^2} = \dfrac{E^2}{4r}\) as well, so the total maximum power drawn from the cell is

\[ P_{\text{total}} = P_{R(\max)} + P_r = \frac{E^2}{4r} + \frac{E^2}{4r} = \frac{E^2}{2r}. \]

(c)(ii)(III) Efficiency of the cell. The efficiency is the ratio of the useful power delivered to the external load to the total power produced by the cell:

\[ \eta = \frac{\text{power in } R}{\text{total power}} = \frac{I^2 R}{I^2 (R + r)} = \frac{R}{R + r} \times 100\% = \frac{V}{V_o} \times 100\%. \]

In particular, at the maximum-power-transfer condition \(R = r\), the efficiency is only \(\dfrac{r}{r+r}\times100\% = 50\%\): half of the energy is wasted inside the cell when the greatest power is transferred to the load.