A battery of emf `1.4 V` and internal resistance `2 Omega` is connected to a resistor of `100 omega` resistance through an ammeter. This resistance of the ammeter is `4//3 Omega`. A voltmeter has also been connected to find the potential difference across the resistor.
a. Draw the circuit diagram.
b. The ammeter reads `0.02 A`. What is the resistance of the voltmeter?
c. The voltmeter reads `1.1 V`. What is the error in the reading?

Answer

2

Explanation

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