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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.`
7 years ago

105 Points
```										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/3Hence 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
All the best.
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Suryakanth –IITB
```
7 years ago
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