2022 A/L Physics Past Paper · Question 41 (English Medium)

Answer: (5) $I_a = I_b = I_c$

What the question asks

What we know:

Three circuits are built from 12 identical resistors $R$ each. Each is joined to the same battery, with emf $E$ and internal resistance $r$.

Circuit (a): the 12 resistors are the 12 edges of a cube. The battery is joined at two opposite corners of the cube.

Circuit (b): a group of 3 resistors, then a group of 6, then a group of 3, joined one after the other.

Circuit (c): three resistors on the left, six in the middle, then three on the right.

No numbers are given. So we keep the letters, and later we check with sample values $R = 12\ \Omega$, $E = 22\ \text{V}$ and $r = 1\ \Omega$.

We want: how the three currents compare.

What each letter means:

$R$ ($\Omega$): the resistance of one of the identical resistors.

$E$ ($\text{V}$): the emf of the battery.

$r$ ($\Omega$): the internal resistance of the battery.

$R_\text{eq}$ ($\Omega$): the equivalent resistance of the 12 resistors in one circuit.

$I$ ($\text{A}$): the current that the battery sends round the circuit.

Relevant theory

What is happening:

The battery is the same in each of the three circuits. So any difference between the currents must come from the resistance of the 12 resistors.

We find the equivalent resistance of each circuit and compare them.

Some of the joints in these circuits are joined to the rest of the circuit in exactly the same way. If swapping such joints leaves the circuit unchanged, they are at the same potential. We may then treat them as one joint.

In circuit (a), look from the corner where the battery is joined. Its three neighbouring corners are alike. The three corners next to the far corner are alike too.

In circuit (c), the three joints after the three left-hand resistors are alike. So are the three joints just before the three right-hand resistors.

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.

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

Resistors joined side by side share the same potential difference. Their currents add. So $n$ identical resistors $R$ side by side have the equivalent resistance $\dfrac{R}{n}$.

Resistors joined one after the other carry the same current. Their resistances add.

For example, a road from a village to a town with three equal bridges side by side across a river. Cars can use any of the three bridges, so the crossing is three times as easy as with one bridge. Then the road goes through six equal lanes side by side, and then three bridges again.

The trip is three stretches one after the other. The difficulty of the trip is the sum of the difficulties of the stretches.

Reasoning and calculation

$\dfrac{R}{3} + \dfrac{R}{6} + \dfrac{R}{3} = \dfrac{2R + R + 2R}{6}$, and each term has the unit $\Omega$, so the sum is a resistance in $\Omega$.

Circuit (a): the cube.

The battery is joined at one corner and at the opposite corner.

The three resistors that leave the first corner are alike, so they carry equal currents and their far ends are at one potential. They are three resistors side by side: $\dfrac{R}{3}$.

From those three corners, six resistors lead to the three corners next to the far corner. Each of the three corners gets two. These six are side by side between two sets of equal potential: $\dfrac{R}{6}$.

The last three resistors lead to the far corner: $\dfrac{R}{3}$.

The total is $\dfrac{5R}{6}$.

Circuit (b): groups of 3, 6 and 3 resistors, one after the other. The total is $\dfrac{R}{3} + \dfrac{R}{6} + \dfrac{R}{3} = \dfrac{5R}{6}$.

Circuit (c): three resistors on the left, each followed by a pair of resistors. Six resistors in the middle lead to three more on the right. In the same way the total is $\dfrac{R}{3} + \dfrac{R}{6} + \dfrac{R}{3} = \dfrac{5R}{6}$.

$10$

$R_\text{eq} = \dfrac{5 \times 12}{6}$

$2$

$I = \dfrac{E}{R_\text{eq} + r} = \dfrac{22}{10 + 1}$

Required equations

$R_1 = \dfrac{R}{3}$

$R_2 = \dfrac{R}{6}$

$R_3 = \dfrac{R}{3}$

$R_\text{eq} = \dfrac{R}{3} + \dfrac{R}{6} + \dfrac{R}{3} = \dfrac{5R}{6}$

Reading the options

The three circuits have the same resistance, $\dfrac{5R}{6}$, and the same battery. So the current is the same:

$$I = \dfrac{E}{\dfrac{5R}{6} + r}$$

So $I_a = I_b = I_c$. That is option (5).

Checks

$e / (5 * rr / 6 + ri)$

Does it make sense?

The shortest way through a circuit crosses three resistors one after the other, which would be $3R$. But there are many ways side by side, so the total is much smaller. $\dfrac{5R}{6}$ is less than $R$.

The same 12 resistors can give different totals, for example $12R$ if they were joined in a row. These three circuits are joined so that they give the same total, so the current is the same.

Conclusion

$I_a = I_b = I_c$

The three circuits have the same resistance, $\dfrac{5R}{6}$.

So $I_a = I_b = I_c$.

The answer is (5).

Find the equivalent resistance of each circuit and compare. Resistors one after the other add. For $n$ identical resistors side by side, divide $R$ by $n$.

If swapping some joints leaves the circuit unchanged, those joints are at the same potential and can be treated as one.

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