Electric circuit with resistors in series

[thereddevils]

Homework Statement

A battery , ammeter with resistance , 1 ohm and a resistor are connected in series .A voltmeter of 500 ohm is connected parallel to the resistor . The actual resistance of R is 100 ohm , what is the ratio of (the resistance R determined)/(the actual resistance of R) ?

Homework Equations

The Attempt at a Solution

Let the emf of the battery be E and resistance of resistor , R

pd across R = [R/(1+R)]E

I(500)= [R/(1+R)]E

and that leads me to nowhere .

 

[rl.bhat]

When the voltmeter is connected across the unknown resistance, what is the equivalent resistance R' of the parallel combination?
Then find PD across R' and the ratio of R'/R.

 

[thereddevils]

When the voltmeter is connected across the unknown resistance, what is the equivalent resistance R' of the parallel combination?
Then find PD across R' and the ratio of R'/R.

ok , the effective resistance , R'=500R/(500+R) and the potential difference across the unknown resistor is

[500R/(501R+500)]E

but that's not the resistance .

 

[skeptic2]

The effective resistance R' is 500R/(550+R) but the potential difference is ER'/(R'+1). (Don't forget the ammeter resistance.)

So now you know the potential difference across R' and the current through it. Can you calculate the determined resistance?

 

[thereddevils]

The effective resistance R' is 500R/(550+R) but the potential difference is ER'/(R'+1). (Don't forget the ammeter resistance.)

So now you know the potential difference across R' and the current through it. Can you calculate the determined resistance?

For the effective resistance , i got 500R/(500+R) , i don see where is the another 50 coming ..

and the pd across the uknown resistor ,

using the concept of potential divider , [500R/(500+R)]/[(500R/(500+R))+1]

=500R/(501R+500)

maybe i am wrong here , could yuo point me to my mistake ? and also i do not know the current flowing through the unkown resistor .

 

[rl.bhat]

If you do the experiment to determine the resistance , you use an ammeter of very small resistance and a voltmeter of very high resistance. Ammeter measures the current and voltmeter measures the voltage across the resistance. Then R = V/I. This is the actual resistance. If the voltmeter is of lower resistance, then the measured resistance is smaller than the actual resistance.
But I can't understand what you mean by
what is the ratio of (the resistance R determined)/(the actual resistance of R) ?

Probably the given data is not sufficient to determine the above ratio.