2023 A/L Physics Past Paper · Question 40 (English Medium)

Answer: (2) $1.0\,\text{A}$

What the question asks

We want: the current $I$ through the cell.

What each letter means:

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

$V$ ($\text{V}$): the voltage of the cell.

$R$ ($\Omega$): the resistance of the whole circuit that the cell sees.

$P$, $Q$, $O$, $S$, $T$, $N$: the joints of the network. $P$ is the joint at the far end of the $2\,\Omega$ resistor next to the cell. $Q$ is the top right corner and $O$ is the middle joint. $S$ and $T$ are the lower left and lower right corners. $N$ is the bottom wire that goes back to the cell.

Given information

What we know:

The cell has $V = 6.5\,\text{V}$.

The internal resistance of the cell is negligible, so the whole $6.5\,\text{V}$ is across the circuit.

The resistors are $2$, $1$, $2$, $8$, $6$, $4$, $2$, $10$, $12$ and $4\,\Omega$, as in the figure.

Relevant theory

If a number of resistors are connected so that the same current flows through them, then it is a series connection. The resistance of the single resistor which carries the same current but is maintained across it, the total potential difference of the system is called the equivalent resistance of the system.

If a number of resistors are connected so that they maintain the same potential difference, then it is a parallel connection. The resistance of the single resistance which carries the total current passing through the system but maintaining the same potential difference across it is called the equivalent resistance of the system.

balanced state, no current passes through the branch BD of the circuit and hence the

.'. Wheatstone Bridge principle is satisfied between A and C. Hence no current flows

The first quote says that resistors in series carry the same current. The second says that resistors in parallel have the same potential difference across them. The third says that when a bridge is balanced, no current passes through its middle branch. The fourth shows a network where this happens, so the middle resistor can be taken away.

For example, think of two roads from a town $P$ to a town $N$, with a cross road between them. If the two ends of the cross road are at the same height, nobody walks along it, because there is no slope to push anyone.

In a circuit, the potential is the height. If two joints are at the same potential, no current passes between them, so the resistor joining them can be taken away.

Reasoning and calculation

Read the circuit.

The $2\,\Omega$ resistor next to the cell is in series with the rest. After it comes the joint $P$, and from $P$ the network goes to the bottom wire $N$, which returns to the cell.

So the whole circuit is the $2\,\Omega$ resistor in series with the network between $P$ and $N$.

Look for symmetry.

Between $P$ and $T$, one path goes through $Q$: $1\,\Omega$ and then $2\,\Omega$. Another goes through $O$: $2\,\Omega$ and then $4\,\Omega$. The two ratios are the same, $1 : 2$.

Between $P$ and $N$, one path goes through $S$: $6\,\Omega$ and then $12\,\Omega$. Another goes through $T$: $2\,\Omega$ (we will see this value soon) and then $4\,\Omega$. Again the ratios are the same, $1 : 2$.

Mark the joints at the same potential.

Because the ratios are the same, $Q$ and $O$ are at the same potential. No current passes through the $8\,\Omega$ resistor between them.

In the same way, $S$ and $T$ are at the same potential. No current passes through the $10\,\Omega$ resistor between them.

So we can take both of these resistors away.

$2$

$R_{PT} = \dfrac{3 \times 6}{3 + 6}$

$4.5$

$R_{PN} = \dfrac{18 \times 6}{18 + 6}$

Required equations

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

$R_{\text{parallel}} = \dfrac{R_1 R_2}{R_1 + R_2}$

$R = 2 + R_{PN}$

$I = \dfrac{V}{R}$

$6.5$

$R = 2 + 4.5$

$1.0$

$I = \dfrac{6.5}{6.5}$

$a * b / (a + b)$

$v / (2 + r)$

Reading the options

Now match it with the five options.

(1) $0.5\,\text{A}$ does not match.

(2) $1.0\,\text{A}$ matches.

(3) $1.2\,\text{A}$ does not match.

(4) $1.5\,\text{A}$ does not match.

(5) $2.0\,\text{A}$ does not match.

Checks

Does it make sense?

The cell sees $6.5\,\Omega$. That is more than the $2\,\Omega$ resistor on its own. It is less than the sum of the resistors in a row, because the other paths help the current through.

Put it back: $1.0\,\text{A}$ through $6.5\,\Omega$ gives $6.5\,\text{V}$, the voltage of the cell.

Conclusion

$I = 1.0\,\text{A}$

The current through the cell is $1.0\,\text{A}$.

The answer is (2).

The quotes give series, parallel and the balanced bridge, where no current passes through the middle branch. That the ratios $1 : 2$ and $1 : 2$ make the joints equal in potential follows from the same current in each path, as shown in the text.

Teacher's note

Look for two paths with the same ratio. Their middle joints are at the same potential, so the resistor between them carries no current and can be taken away.

Then collapse the series and parallel parts, and use $I = \dfrac{V}{R}$.

Exam tip