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.

Theory

An ammeter measures the current that flows through it, so the whole current of the circuit has to pass through the meter: break the circuit and put the ammeter into the gap, in series. A series circuit carries the same current everywhere, and the ammeter shows the same before the resistor and after it. An ideal ammeter has no resistance. A real one adds a little to the circuit, and the less it adds, the less it changes the current it measures.

A voltmeter measures the voltage between the two points its sockets touch. Connect them to the ends of the section, in parallel with it, and the circuit stays closed. An ideal voltmeter draws no current. A real one draws a little past the section, and the greater its resistance, the less it changes the voltage it measures.

The "+" socket of a meter goes on the side of the "+" of the source. With the leads swapped, a digital meter shows a minus sign, and the needle of a dial meter swings the wrong way, below zero.

Swapping the two meters gives two typical mistakes:

• an ammeter across a section shorts it out: only the internal resistance of the source and the ammeter itself limit the current, and a real meter blows its fuse;
• a voltmeter in series stops the current almost entirely: it does no harm, but it shows nearly the whole source voltage instead of the voltage across the section.

A meter with more than one terminal has its own range at each. Connect the lead to the largest range first and move to a smaller one only when the reading fits into it: the meter stays safe, and the needle uses more of the scale. The meters on our bench have two sockets each and no range to choose.

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. The second ammeter puts its 0.1 Ω across the 100 Ω resistor, and almost the whole current takes the path through the meter. Only about 0.8 Ω is left in the circuit: the internal resistance of the source, the two ammeters and the cables. For 6 V across 0.8 Ω Ohm's law gives more than 7 A, far more than the source allows: the overload protection switches it off, and the power indicator flashes red. After the extra ammeter is deleted and the source is switched on, the readings are as before.

The voltmeter connected in series shows about 6.00 V, and the ammeter practically zero (less than a microampere). In series the voltmeter and the rest of the circuit share the source voltage in proportion to their resistances: \(U_V = U \cdot R_V / (R_V + R)\). With 10 MΩ against about 100 Ω the voltmeter gets all but a hundred-thousandth of the 6 V, the current is only about 0.6 µA, and the resistor has about 60 µV across it.

Questions

1. How much does the ammeter lower the current in the circuit, and what share of the resistor's current does the voltmeter draw?
2. Why does a voltmeter connected in series show almost the source voltage while the current in the circuit disappears?

Answers

1. Without the ammeter the circuit has the 100 Ω resistor, the 0.5 Ω of the source and the cables, and \(I = U / R\) gives about 59.7 mA. The ammeter adds its 0.1 Ω and brings the current down to the 59.6 mA on its display, by about 0.1%. The voltmeter, in parallel with the resistor, draws about 5.96 V / 10 MΩ ≈ 0.6 µA, one hundred-thousandth of the 59.6 mA through the resistor.
2. In series the voltmeter's 10 MΩ adds to the roughly 100 Ω of the rest of the circuit, 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: the voltages share out in proportion to the resistances.