A battery of EMF 1.4 V and internal resistance `2Omega` is connected to resistor of `100Omega` resistance through an ammeter. The resistance of the ammeter is `(4)/(3)Omega`. A voltmeter is also connected to find potential difference across resistor. The ammeter reacds 0.02 A if resistance of voltmeter is 100x then x is

Answer

2

Explanation

`R_(eq)=(100xxR)/(100+R)+(4)/(3)+2`
`becauseV=iRimplies1.4=0.02xx((100R)/(100+R)+(4)/(3)+2)`
`R=200Omega`