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# 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.

105 Points
10 years ago

Dear rajan,

Ans :- 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 :

Vb = 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

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