2018 A/L Physics Past Paper · Question 24 (English Medium)

Answer: (4) $I_2 = I_3 > I_1$

What the question asks

We want: how $I_1$, $I_2$ and $I_3$ compare.

All three circuits have the same source $V_0$. So we only need the total resistance of each.

Relevant theory

If a number of resistors are connected so that they maintain the same potential difference

Another fact revealed here is that although the same instrument is used, the ammeters should essentially have low resistances while the voltmeters should essentially have high resistances.

This voltmeter has an internal resistance $r$. It is not infinite.

It is joined across the two ends of $R_1$. So it is in parallel with $R_1$.

Some current goes through the voltmeter too.

Required equations

$I = \frac{V_0}{R_{total}}$

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{r} \;\Rightarrow\; R_{eq} = \frac{R_1 r}{R_1 + r}$

$R_{eq} < R_1$

Reading the options

The options compare $I_1$, $I_2$ and $I_3$.

First find the total resistance of each circuit. Then compare.

Reasoning and calculation

All three circuits have the same $V_0$ and the same $R_2$.

So the current depends only on the total resistance.

Only the top part changes.

In (A) it is $R_1$. In (B) it is $R_1$ with the voltmeter. In (C) it is $R_3$.

Circuit (A)

$I_1 = \dfrac{V_0}{R_1 + R_2}$

The total resistance is $R_1 + R_2$.

$$I_1 = \frac{V_0}{R_1 + R_2}$$

Circuit (B)

$I_2 = \dfrac{V_0}{\frac{R_1 r}{R_1 + r} + R_2}$

$R_1$ and the voltmeter together give $\frac{R_1 r}{R_1 + r}$.

$$I_2 = \frac{V_0}{\frac{R_1 r}{R_1 + r} + R_2}$$

Circuit (C)

$I_3 = I_2$

The question says $R_3 = \frac{R_1 r}{R_1 + r}$.

So (C) has the same total resistance as (B).

$$I_3 = \frac{V_0}{R_3 + R_2}$$

(B) and (C) have the same total resistance. So $I_2$ equals $I_3$.

$$\frac{R_1 r}{R_1 + r} < R_1$$

So (B) has less resistance than (A). Less resistance means more current. So $I_2 > I_1$.

I_2 = I_3

I_1

Less resistance means more current.

(B) and (C) have equal resistance. (A) has the most.

$$I_2 = I_3 > I_1$$

Checks

$0.5$

$I_1 = \dfrac{10}{10 + 10}$

$5$

$R_{eq} = \dfrac{10 \times 10}{10 + 10}$

$0.67$

$I_2 = I_3 = \dfrac{10}{5 + 10}$

If $r$ were infinite, the voltmeter would take no current. Then $I_2$ would equal $I_1$.

Any real $r$ makes $I_2$ larger than $I_1$.

Conclusion

$I_2 = I_3 > I_1$

$$I_2 = I_3 > I_1$$

The answer is option (4).

Thinking the voltmeter changes nothing in the circuit.

That is true only if $r$ is infinite.

A real voltmeter is in parallel with the resistor. It lowers the total resistance.