When an ammter of negligible internal resistance is inserted in series with circuit it reads `1 A` When the voltmeter of very large resistance is connected across `X` it reads `1 V`. When the point `A` and `B` are shorted by a conducting wire, the voltmeter meausres `10 V` across the battery. The internal resistance of the battery is equal to

Options

A. zero
B. `0.5 Omega`
C. `0.2 Omega`
D. `0.1 Omega`

Answer

C

Explanation

(c )`(12)/(X + Y + r) = 1 A implies V_(X) = 1 = 1 xx X implies X = 1 Omega`
When `Y` shorted `I = (12)/(1 + r)`
`10 = 12 - Ir implies 10 = 12 - (12)/((1 + r)^(2))`
`implies 10 + 10 r = 12 + 12 - 12 r`
`implies 10 r = 2 implies r = 0.2 Omega`