2021 A/L Physics Past Paper · Question 40 (English Medium)

Answer: (3) $3\ \text{A}$

What the question asks

We want: the current that the battery gives out.

The battery is joined to the points $A$ and $B$. Sixteen resistors, each $1\ \Omega$, stand between them.

The plan: find the one resistance that can stand in for the whole network. Then use Ohm's law to get the current.

Given information

We know:

Sixteen resistors of $1\ \Omega$ each. A battery of $8\ \text{V}$ whose own resistance can be ignored.

What each letter means:

$E$: the e.m.f. of the battery, $8\ \text{V}$.

$R$: the resistance of one resistor, $1\ \Omega$.

$R_T$ ($\Omega$): the total resistance between $A$ and $B$. This is the one resistance that can stand in for the whole network.

$I$ ($\text{A}$): the current from the battery.

Relevant theory

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

the total charge entering a junction of a circuit should be equal to the total charge leaving the junction.

Reasoning and calculation

Look at the picture. Fold it along the line that goes through $A$ and $B$. The top half lands exactly on the bottom half. Every resistor has a twin.

The two halves are the same, so they carry the same current. Everything we find for the top route is true for the bottom route.

One more thing to notice. The long wire from the left tip runs straight across to the middle of the two upright resistors. Plain wire has no resistance, so the left tip and that middle point are the same point. The same is true on the right.

So the slanting resistor and the upright resistor next to each tip both join the same two points. They are side by side. Resistors side by side are said to be in parallel.

What do resistors side by side do? At a junction, the current splits into parts, and the parts add up to the whole. Each resistor has the same voltage $V$ across it, so each carries $\dfrac{V}{R}$. For $n$ equal resistors the total current is $n\dfrac{V}{R}$. So the group acts like one resistor $\dfrac{R}{n}$.

$0.5$

$R_{left} = \dfrac{1}{2}$

$0.33$

$R_{top} = \dfrac{1}{3}$

$R_{right} = \dfrac{1}{2}$

$1.33$

$R_{route} = 0.5 + \dfrac{1}{3} + 0.5$

$0.67$

$R_{middle} = \dfrac{1.33}{2}$

Now the whole path from $A$ to $B$. It goes through the resistor next to $A$, then the middle part, then the resistor next to $B$. These three are in a line, one after the other. That is called series, and resistors in series add.

$2.67$

$R_T = 1 + \dfrac{2}{3} + 1$

$3$

$I = \dfrac{E}{R_T} = \dfrac{8}{8/3}$

Required equations

$V = IR$

$R_n = \dfrac{R}{n}$

$R_T = R_a + R_b + R_c$

$I = \dfrac{E}{R_T}$

Reading the options

The current is $3\ \text{A}$. This is option (3).

Option (2) puts three side-by-side resistors in a line. Option (4) forgets the middle network. Option (5) is impossible, because the two end resistors alone give $2\ \Omega$. Option (1) adds too many resistors in a line.

Checks

Does it make sense?

The total, $\dfrac{8}{3}\ \Omega$ or $2.67\ \Omega$, is a little more than the $2\ \Omega$ of the two end resistors. The middle network is mostly side-by-side resistors, so it adds only a little.

So the current must be a little less than $\dfrac{8}{2} = 4\ \text{A}$. And $3\ \text{A}$ is.

$v / (a + (b + c + b) / 2 + a)$

Conclusion

$I = \dfrac{E}{R_T} = \dfrac{8}{8/3} = 3\ \text{A}$

Each tip has two resistors side by side, $0.5\ \Omega$. The top has three side by side, $\dfrac{1}{3}\ \Omega$. One route is $\dfrac{4}{3}\ \Omega$, and the two equal routes together are $\dfrac{2}{3}\ \Omega$.

With the two end resistors, $R_T = 1 + \dfrac{2}{3} + 1 = \dfrac{8}{3}\ \Omega$. So $I = \dfrac{8}{8/3} = 3\ \text{A}$. The answer is (3).

Look for a mirror line first. Two identical halves side by side are easy: one route, then halve it.

Find the points that a plain wire joins. They are one point. Then use "side by side" and "in a line" step by step.

Exam tip