# Voltmeter in series with a component?

I always see voltmeters connected in parallel with a component, but what happens if you connect it in series? I think it's due to the high resistance of a voltmeter but I don't really see why this would be a problem: It just means that the current going through the circuit is shifted down a level, but all things remain the same right? (Sorry if this is wrong, I just started learning circuits.)

Thanks

Note: I already saw this but I didn't understand the explanations there.

## 2 Answers

I think it's due to the high resistance of a voltmeter but I don't really see why this would be a problem: It just means that the current going through the circuit is shifted down a level, but all things remain the same right?

Ideally, a voltmeter should have such a high resistance that it's equivalent to an open circuit.

This means that when connected to some device in series, essentially no current will flow through it, and no current will reach the device being tested. Usually that means the device won't operate in a normal way.

• I see, so the resistance of a voltmeter is really high. Is this why it is sometimes said to have infinite resistance? Also what is an open circuit? Thanks! – user45220 May 11 '15 at 15:59
• An ideal voltmeter has infinite resistance. Real voltmeters typically have 1 to 10 megohms equivalent resistance. – The Photon May 11 '15 at 15:59
• Is an open circuit when it acts like an insulator? No charge flows between? Thanks – user45220 May 11 '15 at 16:02
• Yes, an open circuit is the same as infinite resistance. – The Photon May 11 '15 at 16:08

... but what happens if you connect it in series?

Consider a circuit which has a 3-V (ideal) battery connected in series with a 100-$\Omega$ and a 200-$\Omega$ resistor (series resistors). The voltage across the 100-$\Omega$ resistor will be 1 V, and across the 200-$\Omega$, 2 V.

Next, connect a 1 M$\Omega$ resistor (a voltmeter) in parallel with the the 100-$\Omega$ resistor. The voltage across both of these resistors will be approximately 1 V because they are in parallel and must have the same voltage as each other, and the effective resistance of this parallel combination has changed to 99.99 $\Omega$ -- not much. The voltage across the 200-$\Omega$ resistor is still very close to 2 V.

Finally, change the 1 M$\Omega$ resistor to be in series with the 100-$\Omega$ and 200-$\Omega$ resistors. Now the 3 V will be divided between the series resistors, because the total across series resistances must add to the total of the battery in this circuit. Larger resistances will have the larger voltages, so the voltmeter will read slightly less than 3 V, but the voltage across the 100-$\Omega$ resistor is only $3\times 10^{-4}$ V and across the 200-$\Omega$ resistor is $6\times 10^{-4}$ V. You just dramatically changed how the circuit behaved.