In previous chapters, we learned that ammeters are not ideal instruments. When you insert an ammeter, you actually introduce an additional resistance in series with the branch in which you are measuring the current. Generally, this is not a serious problem, but it can have a troubling effect on your readings, so it is important to be aware of it.
Voltmeters also have an internal resistance that appears between the two terminals of interest when a measurement is being made. While an ammeter places an additional resistance in series with the branch of interest, a voltmeter places an additional resistance across the element, as shown in Fig. 1. Since it appears in parallel with the element of interest, the ideal level for the internal resistance of a voltmeter would be infinite ohms, just as zero ohms would be ideal for an ammeter. Unfortunately, the internal resistance of any voltmeter is not infinite and changes from one type of meter to another.

Most digital meters have a fixed internal resistance level in the megaohm range that remains the same for all its scales. For example, the meter in Fig. 1 has the typical level of 11 MΩ for its internal resistance, no matter which voltage scale is used. When the meter is placed across the 10 kΩ resistor, the total resistance of the combination is $$R_T = 10kΩ || 11MΩ$$ $$= {(10^4)(11 \times 10^6) \over (10^4)+(11 \times 10^6)}=9.99kΩ$$ and the behavior of the network is not seriously affected. The result, therefore, is that most digital voltmeters can be used in circuits with resistances up to the high-kilohm range without concern for the effect of the internal resistance on the reading.
However, if the resistances are in the megohm range, you should investigate the effect of the internal resistance.

An analog VOM is a different matter, however, because the internal resistance levels are much lower and the internal resistance levels are a function of the scale used. If a VOM on the 2.5 V scale were placed across the 10 kΩ resistor in Fig. 1, the internal resistance might be 50 kΩ, resulting in a combined resistance of $$R_T=10kΩ || 50kΩ$$ $$={(10^4 Ω)(50 \times 10^3 Ω) \over (10^4 Ω)+(50 \times 10^3 Ω)} =8.33kΩ$$ and the behavior of the network would be affected because the 10 kΩ resistor would appear as an 8.33 kΩresistor.
To determine the resistance $R_m$ of any scale of a VOM, simply multiply the maximum voltage of the chosen scale by the ohm/volt (Ω/V) rating normally appearing at the bottom of the face of the meter. That is, $$R_m (VOM) = \text{(scale)(Ω/V rating)}$$ For a typical Ω/V rating of 20,000, the 2.5 V scale would have an internal resistance of $$(2.5V)(20,000Ω/V)=50kΩ$$ whereas for the 100 V scale, the internal resistance of the VOM would be $$(100V)(20,000Ω/V)=2MΩ$$ and for the 250 V scale, $$(250V)(20,000Ω/V)=5MΩ$$
b. When the meter is connected as shown in Fig. 2 (b), a complete circuit has been established, and current can pass through the circuit. The voltmeter reading can be determined using the voltage divider rule as follows: $$V_{ab}=(11MΩ)(20V)(11MΩ+1MΩ)=18.33V$$ and the reading is affected somewhat.
c. For the VOM, the internal resistance of the meter is $$R_m=(100V)(20,000Ω/V)=2MΩ$$ and $$Vab = {(2 MΩ)(20 V) \over (2 MΩ + 1 MΩ)} = 13.33 V$$ which is considerably below the desired level of 20 V.