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. ` I = 1A`
r is the internal resistance of the battery.
`12= (X+Y+r)(1)`
Also `1 = (X)(1) or X = 1Omega`
Voltage across X (when A and B are shorted)
`10 = (12/(X+r)) X or 10 = (12)/(1+r) or 1 +r = 6/5`
or `r = 1/5 Omega` .