Physics 9702 · Paper 2

Measuring E.m.f. & r

A battery's e.m.f. cannot be read straight off a meter: as soon as it supplies current, every reading drops below E.
The method is indirect. Several current and voltage readings fall on one straight line, and that line gives both E and r.

The measuring circuit

Change the variable resistor, and read V and I again. Each setting gives one pair of numbers, and every pair lies on the same straight line.

The voltmeter is connected across the battery's terminals, so it reads the terminal p.d. V.
The ammeter is connected in the loop, so it reads the current I.
The variable resistor is the part you change: each new setting changes the current, giving a new pair of readings (I, V).

The straight line that gives both numbers

Plot V against I. $V = E - Ir$ has the shape $y = c + mx$ : a straight line with $c = E$ and $m = -r$ .

Plot terminal p.d. against current and the battery gives you both numbers: the intercept is E, the gradient is minus r.

Intercept on the V axis = e.m.f. E: at zero current there are no lost volts, so the terminals show the full e.m.f.
Gradient = −r: each extra ampere of current means another r volts are lost inside.

Your turn— tap to reveal the worked answer (9702/11/M/J/25 Q37)

A cell drives a variable resistor. The graph of p.d. across the resistor against current is a straight line: it meets the p.d. axis at 1.6 V and falls to zero at a current of 80 mA. What is the internal resistance of the cell?

Answer: the size of the gradient is $r = \dfrac{1.6}{0.080} = 20 \; \Omega$ . Check the unit: the current axis is in mA. (9702/11/M/J/25 Q37)

Drag the external resistance below and watch the point (I, V) slide along that exact line. Small R means big current, and the battery wastes most of its e.m.f. on itself.

useful: V = 4.80 Vlost inside: I r = 1.20 V

external R4.0 Ω

I = E / (R + r) = 1.20 A·terminal p.d. = 4.80 V of the 6.00 V label

Worked example

No graph given: two readings are enough

A cell drives a current of 0.625 A through a 2.00 Ω resistor. With a second 2.00 Ω resistor added in series, the current falls to 0.341 A. Find r and E.

Write $E = I(R + r)$ twice: $E = 0.625\,(2.00 + r)$ and $E = 0.341\,(4.00 + r)$ .
Same E, so equate: $0.625\,(2.00 + r) = 0.341\,(4.00 + r)$ .
$1.25 + 0.625\,r = 1.364 + 0.341\,r$ , so $0.284\,r = 0.114$ and $r = 0.40 \; \Omega$ .
Put r back into the first equation: $E = 0.625 \times 2.40 = 1.5$ V.

Answer

r \approx 0.40 \; \Omega, \; E \approx 1.5 \text{ V}

This is the same physics as the graph: two (I, V) points are enough to find the straight line, and the line gives E and r.

The biggest current a battery can give

Join the two terminals with almost no resistance (a ) and the only thing left to limit the current is r itself. The largest possible current is $I_{\max} = E/r$ .
A 3.0 V battery with r of 1.0 Ω can never push more than 3.0 A, no matter what you connect. All that energy turns into heat inside the battery, which is why a short circuit is dangerous.

Worked example

Why headlights dim when the engine starts

A car battery has e.m.f. 12 V and internal resistance 0.04 Ω. The starter motor draws 100 A. Each headlamp is rated 12 V, 36 W. Find the terminal p.d. while starting, and what happens to the headlamp power.

Lost volts: $Ir = 100 \times 0.04 = 4.0$ V, so $V = 12 - 4.0 = 8.0$ V.
Headlamp resistance from its rating: $R = \dfrac{V^2}{P} = \dfrac{12^2}{36} = 4.0 \; \Omega$ .
At 8.0 V: $P = \dfrac{V^2}{R} = \dfrac{8.0^2}{4.0} = 16$ W. Less than half of the normal 36 W, so the lamps clearly become dimmer.

Answer

V = 8.0 \text{ V}, \; P \approx 16 \text{ W}

The bigger the current you demand, the less voltage the battery can give you. That single sentence is the answer template for almost every internal-resistance explain question.

Power and efficiency of a source

Total power made by the chemical reactions: $P_{\text{total}} = EI$ .
Useful power, delivered to the outside circuit: $P_{\text{useful}} = VI$ .
The rest, $I^2 r$ , heats the battery. So the efficiency is $\dfrac{VI}{EI} = \dfrac{V}{E} = \dfrac{R}{R+r}$ .
In the warm-up on the last page (E of 3.0 V, terminal p.d. of 2.75 V), the efficiency is $2.75/3.0 \approx 92\%$ . The other 8% became heat inside the battery.

The load that takes the most power

Make R very small and almost everything is wasted on r. Make R very large and the current is tiny, so the power is tiny too. The useful power is largest in between, exactly when $R = r$ . This result is not on the syllabus, but it explains why engineers try to “match” the resistance of a speaker or an antenna to the source that drives it.

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Measuring E.m.f. & r

A battery's e.m.f. cannot be read straight off a meter: as soon as it supplies current, every reading drops below E.
The method is indirect. Several current and voltage readings fall on one straight line, and that line gives both E and r.

The measuring circuit

Change the variable resistor, and read V and I again. Each setting gives one pair of numbers, and every pair lies on the same straight line.

The voltmeter is connected across the battery's terminals, so it reads the terminal p.d. V.
The ammeter is connected in the loop, so it reads the current I.
The variable resistor is the part you change: each new setting changes the current, giving a new pair of readings (I, V).

The straight line that gives both numbers

Plot V against I. $V = E - Ir$ has the shape $y = c + mx$ : a straight line with $c = E$ and $m = -r$ .

Plot terminal p.d. against current and the battery gives you both numbers: the intercept is E, the gradient is minus r.

Intercept on the V axis = e.m.f. E: at zero current there are no lost volts, so the terminals show the full e.m.f.
Gradient = −r: each extra ampere of current means another r volts are lost inside.

Your turn— tap to reveal the worked answer (9702/11/M/J/25 Q37)

Answer: the size of the gradient is $r = \dfrac{1.6}{0.080} = 20 \; \Omega$ . Check the unit: the current axis is in mA. (9702/11/M/J/25 Q37)

Drag the external resistance below and watch the point (I, V) slide along that exact line. Small R means big current, and the battery wastes most of its e.m.f. on itself.

useful: V = 4.80 Vlost inside: I r = 1.20 V

external R4.0 Ω

I = E / (R + r) = 1.20 A·terminal p.d. = 4.80 V of the 6.00 V label

Worked example

No graph given: two readings are enough

A cell drives a current of 0.625 A through a 2.00 Ω resistor. With a second 2.00 Ω resistor added in series, the current falls to 0.341 A. Find r and E.

Write $E = I(R + r)$ twice: $E = 0.625\,(2.00 + r)$ and $E = 0.341\,(4.00 + r)$ .
Same E, so equate: $0.625\,(2.00 + r) = 0.341\,(4.00 + r)$ .
$1.25 + 0.625\,r = 1.364 + 0.341\,r$ , so $0.284\,r = 0.114$ and $r = 0.40 \; \Omega$ .
Put r back into the first equation: $E = 0.625 \times 2.40 = 1.5$ V.

Answer

r \approx 0.40 \; \Omega, \; E \approx 1.5 \text{ V}

This is the same physics as the graph: two (I, V) points are enough to find the straight line, and the line gives E and r.

The biggest current a battery can give

Join the two terminals with almost no resistance (a ) and the only thing left to limit the current is r itself. The largest possible current is $I_{\max} = E/r$ .
A 3.0 V battery with r of 1.0 Ω can never push more than 3.0 A, no matter what you connect. All that energy turns into heat inside the battery, which is why a short circuit is dangerous.

Worked example

Why headlights dim when the engine starts

Lost volts: $Ir = 100 \times 0.04 = 4.0$ V, so $V = 12 - 4.0 = 8.0$ V.
Headlamp resistance from its rating: $R = \dfrac{V^2}{P} = \dfrac{12^2}{36} = 4.0 \; \Omega$ .
At 8.0 V: $P = \dfrac{V^2}{R} = \dfrac{8.0^2}{4.0} = 16$ W. Less than half of the normal 36 W, so the lamps clearly become dimmer.

Answer

V = 8.0 \text{ V}, \; P \approx 16 \text{ W}

The bigger the current you demand, the less voltage the battery can give you. That single sentence is the answer template for almost every internal-resistance explain question.

Power and efficiency of a source

Total power made by the chemical reactions: $P_{\text{total}} = EI$ .
Useful power, delivered to the outside circuit: $P_{\text{useful}} = VI$ .
The rest, $I^2 r$ , heats the battery. So the efficiency is $\dfrac{VI}{EI} = \dfrac{V}{E} = \dfrac{R}{R+r}$ .
In the warm-up on the last page (E of 3.0 V, terminal p.d. of 2.75 V), the efficiency is $2.75/3.0 \approx 92\%$ . The other 8% became heat inside the battery.

The load that takes the most power