In the given circuit, an ideal voltmeter connected across the `10 Omega` resistance reads 2 V.The internal resistance r, of each cell is :
a) 1Ω b) 0.5Ω c) 1.5Ω d) 0Ω

Options

A. `1Omega`
B. `0.5Omega`
C. `0Omega`
D. `1.5Omega`

Explanation

Current through battery
`i=(1.5+1.5)/((15 times 10)/(15+10)+2+2r)) therefore` Current through `10Omega` resistor
`i_(10 Omega) =15/(15+10) times i=3/5 times I therefore` voltage across `10Omega` resistor `=(3/5 i) times 10=2` (Given)
`implies 3/5 times (3 times 10)/((6+2+2r))=2 implies (8+2r)=90/10=9 implies r=1/2=0.5Omega`