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rajan jha Grade: 12
        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.
6 years ago

Answers : (1)

suryakanth AskiitiansExpert-IITB
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/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


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.


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Askiitians Expert


Suryakanth –IITB

6 years ago
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