When an ammeter of negligible internal resistance is inserted in series with circuit, it reads `1 A`. When a voltmeter of very large resistance is connected across `R_(1)`, it reads `3 V`. But when the points `A` and `B` are short-circuited by a conducting wire, then the voltmetre measures `10.5 V` across the battery. The internal resistance of the battery is equal to

Options

A. `(3)/(7) Omega`
B. `5 Omega`
C. `3 Omega`
D. none of these

Answer

A

Explanation

`3 = IR_(1)` or `3 = 1xx R_(1)` or `R_(1) = 3 Omega`
When points `A` and `B` are connected by a conducting wire, `R_(2)` is short-circuited. Therefore,
`10.5 = I'R_(1)` or `10.5 = I' xx 3`
or `I' = (10.5)/(3) = 3.5 A`
But `10.5 = E - I' r` or `10.5 = 12 - 3. 5r`
`:. r = (1.5)/(3.5) = (3)/(7) Omega`