Guest

A voltmeter of resistance Rv and ammeter of resistance Ra are connected in series across an ideal cell of emf E. When a resistance R is connected parallel to the voltmeter, the reading of the ammeter increases to three times while that of voltmeter reduces to one-third. Find i) Rv:R ii) Ra:R

A voltmeter of resistance Rv and ammeter of resistance Ra are connected in series across an ideal cell of emf E. When a resistance R is connected parallel to the voltmeter, the reading of the ammeter increases to three times while that of voltmeter reduces to one-third.
Find
i) Rv:R
ii) Ra:R

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
Dear student
 
I have considered Rv = R1 and Ra = R2

Case I :
Now the potetial across the battery is given by

Vb = I(R1 + R2).(where I is the current flowing in the circut)
Case II :

Given that the ammeter reading increases 3 times and Hence the net current 3I is drawn from the battery.
Current in voltmeter is reduced to one third,Hence current in voltmeter is I/3
Hence the  Remaining I-I/3 passes through R.

The current through R = 8I/3

VC - VD the potential difference across the voltmeter and the resistance should be same

Hence IR1/3 = 8IR/3

=> R1 = 8R.
Applying Kirchoff's Voltage Rule in ABFGA :

 V= 3IR2 + IR1/3  = I(R1 + R2)   (The battery potentail is same in both the cases)

(R1 + R2) = 3R2+R1/3

=> R2 = R1/3

=> R2 = 8R/3

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free