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Electric circuit with resistors in series

  • #1
438
0

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 .
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
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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.
 
  • #3
438
0


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 .
 
  • #4
1,758
57


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?
 
  • #5
438
0


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 .
 
  • #6
rl.bhat
Homework Helper
4,433
5


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.
 

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