A voltmeter of resistance R1 and an ammeter of resistance R2 are connected in series across a battery of negligible internal resistance . When a resistance R is connected in parrallel to voltmeter , reading of ammeter increases 3 times while that of voltmeter reduces one third. Find R1 and R2 in terms of R.

Dear rajan,

Ans :- Case I :


Now the potetial across the battery is given by

V_b = 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

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

(R₁ + R₂) = 3R₂+R₁/3

=> R_{2 =} R₁/3

=> R_{2 =} 8R/3

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

Askiitians Expert

Suryakanth –IITB