Using ammeter to measure voltage

Suppose you have the following circuit:

Where R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, R4 = 4 kΩ, R5 = 5 kΩ, Vs = 5V. The voltage in the branch with R5 is then 5 I where I over R5 is .064A.

Suppose you put a voltmeter to measure Vab which is the voltage from node Va to node Vb. This would just be the voltage in the branch: 5 * .064, if im not mistaken.

But what if you put an ammeter in to find Vab instead. Since ammeters have low resistance, under ideal conditions, would the voltage measure 0?

• "But what if you put an ammeter in to find Vab instead." What do you mean by "put an ammeter in". In where? Be specific. Oct 18, 2019 at 21:38
• In series with R5 Oct 18, 2019 at 22:18
• 1- There is terminology mistake, there is no such expression "voltage in the branch", instead, for the voltage we say "voltage across R5" ..... 2- If you put ammeter series with R5, you will measure the current, and multiply it by R5 value to get the voltage across R5, If you put it parallel to R5 you will measure the current as if R5 is replaced by zero resistance Oct 19, 2020 at 14:07

Suppose you put a voltmeter to measure Vab which is the voltage from node Va to node Vb. This would just be the voltage in the branch: 5 * .064, if im not mistaken.

Not sure what 5 * .064 means. I think you mean the voltmeter would measure the voltage $$V_{ab}$$. If that's the case, yes it would.

But what if you put an ammeter in to find $$V_{ab}$$ instead. Since ammeters have low resistance, under ideal conditions, would the voltage measure 0?

Based on your clarification that the ammeter is put in series with $$R_5$$, yes it could be used to determine the voltage $$V_{ab}$$, provided that the input impedance of the ammeter is much less than $$R_5$$, just like the input impedance of a voltmeter placed between a and b would have to be much greater than the equivalent impedance between node a and b. But the ammeter would give you the current in $$R_5$$, which would not be a value of 0 volts, unless $$V_{ab}=0$$. It would measure a current equal to

$$I=\frac{V_{ab}}{R_5}$$.

From which you can calculate the voltage $$V_{ab}$$ as

$$V_{ab}=IR_{5}$$

Where $$I$$ = the ammeter reading.

Hope this helps.