2019 A/L Physics Past Paper · Question 45 (English Medium)

Answer: (4) $\frac{E_1r_2r_3 + E_2r_1r_3 + E_3r_1r_2}{r_1r_2 + r_2r_3 + r_1r_3}$

What the question asks

We want:

The potential of the point $P$.

Three cells go from the earthed ring to the same point $P$.

We use $E_k$ and $r_k$.

Relevant theory

Kirchhoff first rule: the currents into a junction add up to zero.

The voltage across the ends of a cell is $V = E - Ir$.

The ring is earthed, so its potential is zero.

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

Required equations

$I_k = \frac{E_k - V_P}{r_k}$

$I_1 + I_2 + I_3 = 0$

Reasoning and calculation

The symbols:

$V_P$ is the potential of $P$.

The ring is at 0 V.

The plus end of each cell is on the $P$ side.

So the current in cell $k$ towards $P$ is $\dfrac{E_k - V_P}{r_k}$.

$\frac{E_1 - V_P}{r_1} + \frac{E_2 - V_P}{r_2} + \frac{E_3 - V_P}{r_3} = 0$

$V_P\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right) = \frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}$

$V_P = \frac{E_1r_2r_3 + E_2r_1r_3 + E_3r_1r_2}{r_1r_2 + r_2r_3 + r_1r_3}$

Reading the options

The form we got:

In the top, each $E_k$ is multiplied by the other two resistances.

The bottom is $r_1r_2 + r_2r_3 + r_1r_3$.

This is exactly option (4).

In (3) the weights are $r_k^2$.

In (5) the bottom is $r_1r_2r_3$.

Neither matches.

Checks

Unit check:

The top has the unit of $[E][r]^2$ and the bottom has $[r]^2$.

So the result is a potential.

Option (5) has the unit of $[E][r]^{-1}$. That is not a potential.

Limit check:

If $r_1 = r_2 = r_3$, then $V_P = \dfrac{E_1 + E_2 + E_3}{3}$. That is right by symmetry.

If $r_1 \to 0$, then $V_P \to E_1$. A cell with no resistance holds $P$ at $E_1$.

$(E1 * r2 * r3 + E2 * r1 * r3 + E3 * r1 * r2) / (r1 * r2 + r2 * r3 + r1 * r3)$

Conclusion

The junction rule gives:

$$V_P = \dfrac{E_1r_2r_3 + E_2r_1r_3 + E_3r_1r_2}{r_1r_2 + r_2r_3 + r_1r_3}$$

So the answer is option (4).