# Why the resistance of a voltmeter has to be infinite?

I know that when resistors are connected in parallel, the potential difference across the resistors remains the same. But when we try to measure a potential difference across a resistor, we make sure that the resistance of the voltmeter be high enough. Why does it matter, in the first place, since the potential difference will be the same across that resister no matter how many resistors in parallel you connect? I know that if the resistor is not high, current will flow through the voltmeter, but I don't understand how the current flow is affecting the voltage across the resistor. Shouldn't The voltage across the resistor remain the same no matter how much current flows through the voltmeter, because the voltmeter and the resistor are in parallel? Any help will be appreciated.

When we use voltmeter in a circuit, we are basically creating a parallel path across a device, which will definitely draw a small amount of current away from the device being tested. This will have an impact on the voltage across that device (because V=IR, and we are reducing I). To minimize this effect, the meter should draw as little current as possible - which happens if its resistance is "very large". We are basically measuring the potential drop, therefore change in current would result in change in potential drop

You would have found the answer anyway by googling.

The current flowing through a resistor does determine the potential difference across it, according to Ohm's Law, $$V=IR$$. If a voltmeter alters the current flowing through a resistor, then its measurement of the voltage drop across the resistor in the circuit without the voltmeter will be wrong.

To put it another way, a finite-resistance voltmeter in parallel with a resistor changes the resistance of that element of the circuit, and therefore changes both the current through it and the voltage drop across it.

For example, suppose a 5-ohm resistor is in a circuit where 1 A of current flows through it, meaning it has a potential difference of 5 V across it. This resistor is in series with another 5-ohm resistor, as well as a power supply of some kind (which is currently supplying 10 V and 1 A). If I place a voltmeter with a 1,000-ohm internal resistance in parallel with this resistor, then the new equivalent resistance is:

$$R_{eq}=\frac{1}{\frac{1}{5}+\frac{1}{1000}}=4.98\;\Omega$$

If the power source is a constant-voltage source, then it continues to supply 10 V of potential difference across the circuit, but, since the equivalent resistance of the whole circuit is now 9.98 ohms instead of 10 ohms, the current supplied will be $$V/R=1.002\;A$$. This means that the resistor you're measuring will now have a voltage drop of $$IR=4.99\; V$$ instead of the 5 V it would have had otherwise. That 10-millivolt difference is caused by placing the voltmeter in the circuit.

If, on the other hand, the power source is a constant-current source, then it continues to supply 1 A of current, which means that the measured voltage drop across the resistor will be $$4.98\;V$$ instead of the 5 V it would have been otherwise.

In both cases, a finite resistance biases your measurement of voltage drop downward.

• Re, "finite-resistance...in parallel with a resistor...changes both the current through it and the voltage drop across it." Counter-example: Resistor, R, is connected in parallel with a constant-voltage source, V. The voltage across the resistor is V, the current is V/R. Adding additional resistors in parallel with the original will not change the voltage, V, and therefore the current through the first resistor will not change either. But in other, more complicated circuits, adding an additional resistance in parallel with R would change both the voltage across and the current through R. Jan 29, 2020 at 21:50
• @SolomonSlow This is why, if you re-read the answer, there are two resistors in series in the original circuit, and the voltmeter is applied to one of them. If that second resistor wasn't there, there wouldn't be a problem. Jan 29, 2020 at 22:52

Suppose you have a simple circuit consisting two resistances, $$R_1$$ and $$R_2$$ connected in series across a constant voltage source. Now suppose that you connect a voltmeter of finite resistance across $$R_2$$. If the voltmeter is correctly calibrated it will indeed read the potential difference (pd) across $$R_2$$. But that pd will be less than what it would have been before the voltmeter was connected! This is because the parallel combination of voltmeter and $$R_2$$ will have less resistance than $$R_2$$ by itself. So connecting the voltmeter will increase the current through $$R_1$$, meaning that a greater pd will be dropped across $$R_1$$, leaving a smaller proportion of the supply voltage across $$R_2$$ (and the voltmeter)!

An example: Supply voltage = 20 V, $$R_1 = 1\ \text{k}\Omega,\ R_2=4\ \text{k}\Omega$$, voltmeter resistance = $$12\ \text{k}\Omega$$. You should find that the pd across $$R_2$$ is 16 V without the voltmeter, but 15 V with the voltmeter.

If $$R_1$$ were not present, then placing the voltmeter in parallel with $$R_2$$ wouldn't affect the pd across $$R_2$$. But note that we're assuming a constant voltage source. Most real voltage sources (such as batteries) have internal resistances (a hidden $$R_1$$).