2018 A/L Physics Past Paper · Question 42 (English Medium)

A resistor network in a box — voltmeter and ammeter

In the circuit shown in figure (a), the box with broken lines contains a resistor network. The 2 V battery has a negligible internal resistance. An ideal voltmeter connected across $ab$ gives a reading of 1 V. When the voltmeter is replaced by an ideal ammeter it indicates 2 mA. The resistor network inside the box with broken lines is

(1) Figure (1)
(2) Figure (2)
(3) Figure (3)
(4) Figure (4)
(5) Figure (5)

Answer and explanation

Answer: (3) Figure (3)

What the question asks

We want the network that gives both readings.

With an ideal voltmeter across $ab$ , the reading is $1$ V.

With an ideal ammeter across $ab$ , the reading is $2$ mA.

Relevant theory

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 means that the electric charges flow without getting collected at any point or a junction of a circuit.

An ideal voltmeter takes no current.

An ideal ammeter has no resistance. It joins $a$ to $b$ .

A resistor with no current has no potential difference across it.

Required equations

$V = IR$

$V_b = \frac{R_2}{R_1 + R_2}\,E$

Reading the options

Test all five networks with both readings.

Reasoning and calculation

The battery gives $E = 2$ V. $c$ is at $+2$ V and $d$ is at $0$ .

Voltmeter: $V_{ab} = 1$ V.

Ammeter: $I = 2$ mA.

With the voltmeter, $b$ must be at $1$ V. So $b$ must join the middle of a potential divider.

With the ammeter, the path may hold only $1$ kΩ.

$R = \frac{2\ \text{V}}{2\ \text{mA}} = 1\ \text{k}\Omega$

Every network joins $c$ to $a$ with a wire.

Only the 1 kΩ resistors are joined in different ways.

Only (3) gives both readings.

(1)

$1$ V ✓, $0.67$ mA ✗

(2)

$2$ V ✗

(3)

$1$ V ✓, $2$ mA ✓

(4)

(5)

$1$

$V_b = \frac{1}{1 + 1} \times 2$

$V_{ab} = 2 - 1$

$0.002$

$I = \frac{2}{1000}$

Checks

$E / R$

In (2), (4) and (5), the resistor at $b$ carries no current. So $b$ is at the potential of $d$ . The voltmeter would read $2$ V.

In (1) the ammeter current is less than $2$ mA.

Only (3) fits both readings.

The voltmeter reading of $1$ V fits.

But with the ammeter, the 1 kΩ at $b$ is also in the path. The current is only about $0.67$ mA.

The 1 kΩ at $b$ carries no current. So $b$ is at the potential of $d$ .

$V_{ab} = 2\ \text{V}$

$b$ is joined to the $d$ line through 1 kΩ. No current flows in it.

Conclusion

Network (3) gives $1$ V with the voltmeter and $2$ mA with the ammeter.

The answer is option (3).

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