2019 A/L Physics Past Paper · Question 49 (English Medium)

Answer: (4) $\frac{V}{R}$

What the question asks

We want:

The total current $I$ through the battery.

The diamond network is joined across the battery $V$.

Given information

The nodes:

$A$ is the left corner, at potential $V$.

$C$ is the right corner, at $0$.

$O$ is the centre, $T$ the top, $B$ the bottom.

On the left inside are $P$ (upper) and $Q$ (lower). On the right are $P^{\prime}$ and $Q^{\prime}$.

The resistors:

$A$–$P$, $A$–$O$ and $A$–$Q$ are each $R$.

$P$–$O$ is $2R$.

$P$–$T$ is $4R$.

$P$–$P^{\prime}$ is $8R$.

The right side and the bottom are the same.

Relevant theory

Junction rule: the currents into a node are the currents out of it.

Symmetry tells us which nodes are at the same potential.

That makes the network simple.

The algebraic sum of the currents flowing across a junction of a circuit is zero

Reasoning and calculation

The shape:

Three paths leave $A$, to $P$, $O$ and $Q$.

Three paths reach $C$, from $P^{\prime}$, $O$ and $Q^{\prime}$.

The top $T$ and the bottom $B$ are joined by $4R$ resistors.

The $8R$ resistors are between $P$ and $P^{\prime}$, and between $Q$ and $Q^{\prime}$.

The symmetry:

Left and right are mirror images, with $V$ and $0$ swapped. So $O$, $T$ and $B$ on the middle line are all at $\dfrac{V}{2}$.

Up and down are the same. So $V_P = V_Q = x$ and $V_{P^{\prime}} = V_{Q^{\prime}} = V - x$.

Nodes at the same potential:

$O$, $T$ and $B$ at $\dfrac{V}{2}$.

$P$ and $Q$ at $x$.

$P^{\prime}$ and $Q^{\prime}$ at $V - x$.

So we have only one unknown, $x$, not six.

What we can leave out:

The potential of $O$, $T$ and $B$ is already known. We need no equation for them.

The equations at $Q$, $P^{\prime}$ and $Q^{\prime}$ copy the one at $P$.

So one junction rule at $P$ is enough.

$8(V - x) + 2\left(\frac{V}{2} - x\right) + \left(V - 2x\right) + 4\left(\frac{V}{2} - x\right) = 0$

$12V - 16x = 0$

$\frac{3V}{4}$

$x = \frac{3V}{4}$

$\frac{V}{4R}$

$\frac{V - 3V/4}{R}$

$\frac{V}{2R}$

$\frac{V - V/2}{R}$

Required equations

$I = \frac{V}{4R} + \frac{V}{4R} + \frac{V}{2R}$

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

Reading the options

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

That is option (4).

The whole network works like one resistor $R$.

Checks

$V / R$

Check:

The current into $P$ is $\dfrac{V}{4R}$.

Out of $P$: to $O$ it is $\dfrac{V/4}{2R} = \dfrac{V}{8R}$.

To $T$ it is $\dfrac{V/4}{4R} = \dfrac{V}{16R}$.

To $P^{\prime}$ it is $\dfrac{V/2}{8R} = \dfrac{V}{16R}$.

The total is $\dfrac{V}{4R}$, so it matches.

Conclusion

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

By symmetry $O$ is at $\dfrac{V}{2}$. $P$ and $Q$ are at $\dfrac{3V}{4}$.

From $A$ the currents are $\dfrac{V}{4R} + \dfrac{V}{4R} + \dfrac{V}{2R} = \dfrac{V}{R}$.

So the answer is option (4).