# Homework Help: Calculating Voltmeter's Resistance

1. May 24, 2012

### ZxcvbnM2000

1. The problem statement, all variables and given/known data

A battery is known to have an emf of 5 V, but when a particular voltmeter is
connected to it the reading is 4.9 V. With the voltmeter disconnected and a load
resistance of 12 Ω connected to the battery, a current of 0.4 A is delivered. What
is the resistance of the voltmeter?

2. Relevant equations

3. The attempt at a solution

Okay first of all i found that the internal resistance of the battery is 0.5 Ω .
since I = E/(R+r) .

But to find the meter's resistance is a bit confusing ...

Since their is a voltage drop of 0.1 V it means that the voltage drop across r and the voltage drop across Rv(voltmeter's resistance) should be equal to 0.1 V right ?

So Vr + Vx =0.1 ?

I have no idea ?! :S

2. May 24, 2012

### Staff: Mentor

In the voltmeter measurement, what is that 0.1V dropping across.... And what important info does knowing that give you...?

3. May 24, 2012

### Staff: Mentor

I think you can assume that the voltage that the voltmeter displays is the voltage across its own resistance.

4. May 24, 2012

### ZxcvbnM2000

So we can say that Vterminal = E - Ir ?

so 4.9 = 5 - Ir <=> 0.1 =Ir so the voltage drop across the battery's internal resistance is 0.1 V so this either means that the voltmeter has zero resistance or they are in parallel.Right ?
But if they are in parallel then all 5 volts should be available right ??!!!

Argh !

5. May 24, 2012

### Staff: Mentor

Sure.
No, it means that the current is such that 0.1V is dropped across the internal resistance. It doesn't determine the voltmeter's resistance. If the leads of the voltmeter are connected across the battery terminals, then the voltmeter resistance must be in series with the internal resistance of the battery.

https://www.physicsforums.com/attachment.php?attachmentid=47633&stc=1&d=1337910519
They are not in parallel.

You have two circuits to consider. One has the voltmeter resistance as a load and the the other has 12 Ω as a load. You're given two resulting "measurements". Both situations have some common components. You should be able to write equations for each situation, each with one unknown value, and solve for the unknowns.