# Electric circuit with resistors in series

## 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) ?

## 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 .

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rl.bhat
Homework Helper

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.

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 thats not the resistance .

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?

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
Homework Helper

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.