2020 A/L Physics Past Paper · Question 33 (English Medium)

Answer: (2) $4\ \Omega$

Given information

Given: a cell of e.m.f. $E$, internal resistance $r$; switch $K$ connects either $R_1 = 8\ \Omega$ or $R_2 = 2\ \Omega$ in series with the cell; in both positions the power dissipated in the resistor is the same.

What the question asks

We must find: the internal resistance $r$.

Relevant theory

The electromotive force of a cell is defined as the amount of form of energy existing in the cell which gets converted into electrical energy when a unit electric charge passes through a circuit which includes the cell.

Aresistance acts against the flow of the current inside the cell and it is called the internal resistance of the cell. This resistance exists in series with the resistance of the external circuit.

E = IR + Ir

P=VI

Required equations

$I = \frac{E}{R + r}$

$P = I^2R = \frac{E^2R}{(R + r)^2}$

$P_8 = P_2$

Applying it to this question

Plan: (1) the current in each position; (2) the power in each resistor; (3) equate the two and solve for $r$. Even without knowing $E$, it cancels because $E^2$ appears on both sides.

Reasoning and calculation

$\frac{8E^2}{(8 + r)^2}$

$I_1 = \frac{E}{8 + r},\quad P_8 = I_1^2 \times 8$

$\frac{2E^2}{(2 + r)^2}$

$I_2 = \frac{E}{2 + r},\quad P_2 = I_2^2 \times 2$

$\frac{8E^2}{(8 + r)^2} = \frac{2E^2}{(2 + r)^2} \;\Rightarrow\; \frac{4}{(8 + r)^2} = \frac{1}{(2 + r)^2}$

$\frac{2}{8 + r} = \frac{1}{2 + r}$

$2(2 + r) = 8 + r \;\Rightarrow\; 4 + 2r = 8 + r$

$4$

$r = 8 - 4$

Comparing with the options: $r = 4\ \Omega$ — option (2). This is not the average of $2\ \Omega$ and $8\ \Omega$; it is the square root of their product.

Checks

$8 * pow(E / (8 + r), 2) - 2 * pow(E / (2 + r), 2)$

Second way (general formula): expanding $\dfrac{R_1}{(R_1 + r)^2} = \dfrac{R_2}{(R_2 + r)^2}$ gives $(R_1 - R_2)(r^2 - R_1R_2) = 0$; since $R_1 \ne R_2$, $r^2 = R_1R_2 = 8 \times 2 = 16$, $r = 4\ \Omega$ ✓. Numeric check: with $E = 12$ V, $I_1 = \dfrac{12}{12} = 1$ A, $I_2 = \dfrac{12}{6} = 2$ A; $1^2 \times 8 = 2^2 \times 2 = 8$ W ✓.

Physical plausibility: $r = 4\ \Omega$ is positive and lies between the two loads. If $r$ were very small, the terminal voltage $\approx E$ for both, and the smaller resistor $2\ \Omega$ would always take the larger power ($\dfrac{E^2}{R}$); if $r$ were very large, the current $\approx \dfrac{E}{r}$ for both, and the larger resistor $8\ \Omega$ would take the larger power ($I^2R$) — somewhere between, they must be equal ✓.

Unit check: $\dfrac{\text{V}^2 \cdot \Omega}{\Omega^2} = \dfrac{\text{V}^2}{\Omega} = \text{W}$ ✓; the unit of $r = \sqrt{R_1R_2}$ is $\Omega$ ✓.

Conclusion

$r = \sqrt{8 \times 2} = 4\ \Omega$

From $\dfrac{8E^2}{(8 + r)^2} = \dfrac{2E^2}{(2 + r)^2}$, $2(2 + r) = 8 + r$: the internal resistance $r = 4\ \Omega$ — the answer is (2).

Ignoring the internal resistance and assuming the same potential difference $E$ appears across both resistors.

The internal resistance is in series with the external resistance: $I = \dfrac{E}{R + r}$, so the terminal potential difference changes with $R$.