2023 H2 Physics Paper 2 Q5 — A Thermistor's Curve and Two Resistors Two Ways

The model, live

I–V curve / currentLine from the origin (R = V/I)Parallel circuit

Part (a)

2 marks

Describe and explain what the thermistor's I–V graph shows about its resistance as the p.d. increases from zero.

Approach

Resistance is $V/I$ at a point, which is the reciprocal of the gradient of the line from the origin to that point (not the tangent). Describe how that changes along the curve, then give the physical reason.

Near the origin the graph is a straight line through the origin, so $V/I$ is constant. Further along, the curve bends upwards: each extra volt gives more extra current, so $V/I$ falls. A larger current heats the thermistor; as its temperature rises, more charge carriers are released, so its resistance decreases.

At low p.d. the resistance is constant (straight line through the origin). As the p.d. increases further, $I$ rises more than in proportion to $V$ , so the resistance $V/I$ decreases. This is because the current heats the thermistor, and at a higher temperature more charge carriers are available for conduction.

Watch

A steeper I–V graph means a smaller resistance. Don't mix it up with a V–I graph, where steeper means larger.

Part (b)(i)

2 marks

For the $220\,\Omega$ and $640\,\Omega$ resistors in series, show that the ratio of power in the $220\,\Omega$ to power in the $640\,\Omega$ is $0.34$ , explaining the working.

Approach

Pick the power formula whose other quantity is the same for both resistors. In series that is the current, so use $P = I^2R$ .

\frac{P_{220}}{P_{640}} = \frac{I^2(220)}{I^2(640)} = \frac{220}{640} = 0.344

In series the same current $I$ flows through both resistors. With $P = I^2R$ , the $I^2$ cancels, so $\dfrac{P_{220}}{P_{640}} = \dfrac{220}{640} = 0.34$ .

Part (b)(ii)

1 mark

The supply is replaced by one with internal resistance. State and explain the effect on the ratio in (b)(i).

No effect. The internal resistance reduces the current, but the two resistors are still in series and carry the same current as each other, so the ratio of powers is still $\dfrac{220}{640} = 0.34$ .

Part (b)(iii)

3 marks

With the original supply (no internal resistance), find the ratio of the supply current in series to the supply current in parallel.

Approach

The supply p.d. $V$ is the same in both circuits, so each current is $V$ divided by the total resistance. The ratio of currents is the inverse ratio of the resistances.

R_\text{series} = 220 + 640 = 860\,\Omega

\frac{1}{R_\text{parallel}} = \frac{1}{220} + \frac{1}{640} \quad\Longrightarrow\quad R_\text{parallel} = 163.7\,\Omega

\frac{I_\text{series}}{I_\text{parallel}} = \frac{V/860}{V/163.7} = \frac{163.7}{860} = 0.190

To check your answer:

Take $V = 12\,\text{V}$ . In series $I = 12/860 = 0.0140\,\text{A}$ . In parallel the branch currents are $12/220 = 0.0545\,\text{A}$ and $12/640 = 0.0188\,\text{A}$ , total $0.0733\,\text{A}$ . The ratio is $0.0140/0.0733 = 0.19$ .

ratio $= 0.19$

Watch

The parallel resistance is smaller than either resistor, so the parallel current is the larger one and the ratio must be less than 1.

Part (b)(iv)

1 mark

The supply is replaced by an a.c. supply, with no internal resistance, whose peak voltage equals the d.c. output voltage. State and explain the effect on the ratio in (b)(iii).

No effect. At every instant the same p.d. is across both circuits and the resistances are unchanged, so each current is still the p.d. divided by the total resistance. The instantaneous, peak and r.m.s. currents are all in the ratio $163.7 : 860 = 0.19$ .

What this question was really testing

Resistance from an I–V graph. $R = V/I$ at the point, and the reason a thermistor's falls: heating releases more charge carriers.
Choosing the right power formula. $I^2R$ when the current is shared (series), $V^2/R$ when the p.d. is shared (parallel).
Ratios that cancel. Changing the supply changes the currents, but it cannot change a ratio fixed only by the resistances.

Ratios that survive a change of supply, and a curve that bends upwards

Part (a)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

Part (b)(iv)

What this question was really testing

This is how we mark, and how we teach.