Question

Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part a a voltmeter with resistance rv is connected across the terminals of a battery of emf e and internal resistance r. find the potential difference vmeter measured by the voltmeter.

Answer 1

The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:  
R = internal resistance r + resistance connected rv 
R = r + rv  
Now find the current:  
V 1= IR 
I = R / V1  
find the voltage at the battery terminal (which is net of internal resistance) using  
V 2= IR  
So the voltage at the terminal is:  
V = V2 - V1  
This is the potential difference vmeter measured by the voltmeter.

Answer 2

Answer:

$\Delta V = \frac{E R_v}{R_v + r}$

Explanation:

A real voltmeter is connected to a non ideal battery of EMF "E"

here the resistance of the voltmeter is given as

$R = R_v$

also the internal resistance of the cell is given as

$R_{in} = r$

now the total resistance of the given combination of cell and the voltmeter is given as

$R = R_{in} + R_v$

$R = r + R_v$

now as per ohm's law the current is given as

$i = \frac{V}{R}$

$i = \frac{E}{R_v + r}$

now potential difference across the voltmeter is given as

$\Delta V = i R_v$

$\Delta V = \frac{E R_v}{R_v + r}$