2023 A/L Physics Past Paper · Question 29 (English Medium)

Answer: (4) $24\,\text{k}\Omega$

What the question asks

We want: the resistance $R$ of one resistor.

What each letter means:

$R$ ($\text{k}\Omega$): the resistance of one resistor.

$R_{AB}$ ($\text{k}\Omega$): the resistance of the whole circuit between $A$ and $B$.

$X$, $Y$, $Z$: the first, second and third upright wires in the figure, counting from the left.

Given information

What we know:

There are ten resistors. Each one has the resistance $R$.

The resistance between the ends $A$ and $B$ is $R_{AB} = 50\,\text{k}\Omega$.

Relevant theory

If a number of resistors are connected so that the same current flows through them, then it is a series connection.

If a number of resistors are connected so that they maintain the same potential difference, then it is a parallel connection.

$R = R_1 + R_2 + R_3$

The first quote says that resistors in series carry the same current. The second quote says that resistors in parallel have the same potential difference across them. The third quote says that in series the resistances add.

For example, think of a road from $A$ to $B$. Where the road has one lane, the traffic crawls. Where it opens into more lanes, the traffic gets through more easily.

Each lane is as hard to drive as the others. So two lanes are twice as easy as one, three lanes are three times as easy, and four lanes are four times as easy.

In a circuit, $n$ equal resistors in parallel are the same as one resistor of $\dfrac{R}{n}$. Each has the same potential difference across it, so each carries the same share of the current, and together they carry $n$ times as much as one.

Reasoning and calculation

Read the circuit from $A$ to $B$.

First there is one resistor on its own. After it come three groups, one after the other. Each group is a set of resistors between two upright wires: first $X$ and $Y$, then $Y$ and $Z$, then $Z$ and $B$.

Look for symmetry.

In each group the resistors are the same, $R$, and they join the same two upright wires. So the same potential difference is across each of them, and each carries the same current. The top and bottom resistors of the first group are mirror copies of each other, and so are those of the other groups.

A wire has no resistance, so a whole upright wire is at one potential. So $X$ is one point of the circuit, and $Y$ and $Z$ are too.

The groups sit between these points one after the other, so we can collapse each group on its own.

$R/2$

$R_{XY} = \dfrac{R}{2}$

$R/3$

$R_{YZ} = \dfrac{R}{3}$

$R/4$

$R_{ZB} = \dfrac{R}{4}$

Required equations

$R_{\text{series}} = R_1 + R_2 + R_3 + \dots$

$R_{\text{parallel}} = \dfrac{R}{n}$

$R_{AB} = R + R_{XY} + R_{YZ} + R_{ZB}$

The resistances are in $\text{k}\Omega$, so we can keep the $\text{k}\Omega$ and write $R$ for the number in $\text{k}\Omega$.

$\dfrac{25}{12}R$

$R_{AB} = R + \dfrac{R}{2} + \dfrac{R}{3} + \dfrac{R}{4}$

$R = 50 \times \dfrac{12}{25}$

$\dfrac{25}{12}R = 50$

$24$

$R = 24$

$a + a / 2 + a / 3 + a / 4$

Reading the options

Now match it with the five options.

(1) $12\,\text{k}\Omega$ gives $25\,\text{k}\Omega$ between $A$ and $B$. It does not match.

(2) $15\,\text{k}\Omega$ gives $31.25\,\text{k}\Omega$. It does not match.

(3) $18\,\text{k}\Omega$ gives $37.5\,\text{k}\Omega$. It does not match.

(4) $24\,\text{k}\Omega$ gives $50\,\text{k}\Omega$. This matches.

(5) $36\,\text{k}\Omega$ gives $75\,\text{k}\Omega$. It does not match.

Checks

Put $R = 24\,\text{k}\Omega$ back and add the parts: $24 + 12 + 8 + 6 = 50\,\text{k}\Omega$. It matches the question.

Another look: each parallel group has a smaller resistance than one resistor alone, so the total must be less than the ten resistors in a row, $10R$. And it must be more than the first resistor $R$ alone, because that one is in the way of the current. $50\,\text{k}\Omega$ is $\dfrac{25}{12} \times 24$, about $2.1$ times $R$. That is between $R$ and $10R$.

Conclusion

$R = 24\,\text{k}\Omega$

Conclusion

The quotes give the idea of series and of parallel, and that series resistances add. That $n$ equal resistors in parallel are worth $\dfrac{R}{n}$ follows from the same potential difference across each of them, as shown in the text.

Teacher's note

Collapse the circuit one group at a time. Resistors that share the same two ends are in parallel. Groups that follow each other are in series.

For $n$ equal resistors in parallel, use $\dfrac{R}{n}$. Then add the series parts.

Exam tip