2021 A/L Physics Past Paper · Question 31 (English Medium)

Answer: (3) $2.0\ \Omega$

Given information

We know:

With key $K$ open, the balance length is $L = 60.0\ \text{cm}$. With key $K$ closed, the balance length is $l = 50.0\ \text{cm}$. The resistor $R$ is $10\ \Omega$.

What each letter means:

$E$ ($\text{V}$): the e.m.f. of the cell (the full push it can give).

$r$ ($\Omega$): the internal resistance of the cell (the cell's own resistance).

$R$: the resistor joined across the cell when $K$ is closed, $10\ \Omega$.

$V$ ($\text{V}$): the voltage across the cell's terminals while current flows.

$I$ ($\text{A}$): the current through the cell and $R$.

$L$ and $l$ ($\text{cm}$): the balance lengths with $K$ open and closed.

$k$ ($\text{V cm}^{-1}$): the voltage that each centimetre of the wire holds.

What the question asks

We want: the internal resistance $r$ of the cell.

The plan: use the two balance lengths to compare the two voltages. Then use the circuit with $R$ to link the voltages to $r$.

Relevant theory

the first step is to keep the cell in open circuit and find the balance length L.

Since the cell is now in closed circuit what is being balanced is a terminal potential difference (V) which is less than the e.m.f.

Since no current passes through the cell in the balanced position, its internal resistance has no effect on its e.m.f..

The current flowing in a conductor is directly proportional to the potential difference between its ends provided temperature and its other physical factors remain constant

Required equations

$E = kL$

$V = kl$

$E = I(R + r)$

$V = IR$

$\dfrac{E}{V} = \dfrac{R + r}{R}$

Applying it to this question

Think of the wire as a long ramp. The voltage drops evenly along it, so each centimetre holds the same small voltage $k$. The longer the balance length, the bigger the voltage it balances.

Think of the cell as a water pump. With $K$ open, no water flows. We feel the pump's full push, $E$.

With $K$ closed, water flows through the $10\ \Omega$ pipe. Some of the push is used up inside the pump itself, on its own narrow part $r$. So the push we get outside, $V$, is smaller.

$\dfrac{E}{V} = \dfrac{kL}{kl} = \dfrac{L}{l}$

$1.2$

$\dfrac{E}{V} = \dfrac{60}{50}$

$\dfrac{R + r}{R} = \dfrac{E}{V}$

$1 + \dfrac{r}{R} = 1.2$

$0.2$

$\dfrac{r}{R} = 0.2$

$2$

$r = 0.2 \times 10$

$\dfrac{r}{R}$ has no unit, because both are resistances. So $r = 0.2 \times R$ is in ohms, the unit of $R$.

Reasoning and calculation

The voltages are in the ratio $60 : 50$. That means the cell loses $\dfrac{1}{6}$ of its push inside itself, since $60 - 50 = 10$ out of $60$.

The part lost inside $r$ is $10$ for every $50$ left across $R$. So $r$ is $\dfrac{10}{50}$ of $R$, which is $0.2 \times 10\ \Omega = 2\ \Omega$. This is option (3).

Checks

$a * (b - c) / c$

Does it make sense?

The balance length with $K$ closed is shorter than with $K$ open. That is right: a cell that is giving current has a smaller voltage at its terminals.

The internal resistance, $2\ \Omega$, is much smaller than $R = 10\ \Omega$. A cell that is working well has a small internal resistance, and the two lengths are close, $60$ and $50$.

Conclusion

$r = R\,\dfrac{L - l}{l} = 2\ \Omega$

The voltages are in the ratio $\dfrac{E}{V} = \dfrac{60}{50} = 1.2$. The circuit gives $\dfrac{R + r}{R} = 1.2$, so $\dfrac{r}{R} = 0.2$.

$r = 0.2 \times 10 = 2\ \Omega$. The answer is (3).

With $K$ open the potentiometer gives $E$. With $K$ closed it gives $V$. Their ratio is the ratio of the two lengths.

Then use $r = R\,\dfrac{L - l}{l}$. Check that the answer comes out in ohms.

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