Connecting an ammeter and a voltmeter

Curricula: KS3, cycle 4, Sek I

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Aim

Learn to connect an ammeter in series and a voltmeter in parallel with a section of the circuit.

The bench

A 6 V source, an ammeter, a 100 Ω resistor; a voltmeter is connected in parallel with the resistor. The meters are in DC mode.

Procedure

1. Write down the readings of the ammeter and the voltmeter.
2. Drag a second ammeter from the elements panel and connect it with cables in parallel with the resistor.
3. Delete the extra ammeter and switch the source on with the power button.
4. Connect the voltmeter into the circuit in series: remove the cable between the ammeter and the resistor and move the voltmeter's cables — the cable from "+" to the output of the ammeter, the cable from "−" to the left socket of the resistor. Write down the readings of both meters.

Expected result

The ammeter shows about 59.6 mA, the voltmeter about 5.96 V. An ammeter connected in parallel has almost no resistance and short-circuits the resistor: only the internal resistance of the source and the two ammeters remain in the circuit. The current exceeds 7 A, the overload protection switches the source off, and the power indicator flashes red. After the extra ammeter is deleted and the source is switched on, the readings are as before.

A voltmeter connected in series shows almost the whole source voltage, about 6.00 V, and the ammeter practically zero (less than a microampere): the voltmeter's resistance of 10 MΩ is a hundred thousand times greater than the resistor's, and it hardly lets any current pass. Meanwhile there is almost no voltage across the resistor: a meter connected in series "takes" all of it.

Questions

1. Why must an ammeter have a low resistance and a voltmeter a high one?
2. Why does a voltmeter connected in series show almost the source voltage while the current in the circuit disappears?

Answers

1. The ammeter is connected in series, so its resistance adds to the circuit and must be small for the meter not to reduce the current it measures: here 0.1 Ω against 100 Ω. The voltmeter is connected in parallel and draws part of the current around the section it measures; with 10 MΩ against 100 Ω that part is about one hundred-thousandth of the current through the resistor.
2. In series the voltmeter's 10 MΩ is added to the 100 Ω resistor, so the current is only \(I = U / R\) ≈ 6 V / 10 MΩ = 0.6 µA. At this current the resistor has only about 60 µV across it, the ammeter and the source even less, and almost the whole 6 V is left across the voltmeter.