When a battery connected across a resistor of `16 Omega`, the voltage across the resistor is 12 V. When the same battery is connected across a resistor of `10 Omega`, voltage across it is 11 V. The internal resistance of the battery (in ohm) is

Options

A. `(10)/(7)`
B. `(20)/(7)`
C. `(25)/(7)`
D. `(30)/(7)`

Answer

B

Explanation

Here, VltE
`lt`E = V + lr
For first case,
`E=12+(12)/(16)r `
For second case,
`E=11+(11)/(10)r
From Eqs. (i) and(ii),we get
`12+(12)/(16)r=11+(11)/10rrArr r==20/7Omega`