Two identical calls send the same current in `2 Omega` resistance, whether connected in series or in parallel. The internal resistance of the cell should be

Options

A. `1 Omega`
B. `2 Omega`
C. `1/2Omega`
D. `2.5 Omega`

Answer

B

Explanation

In series, `i_(1) =(2E)/(2 + 2r)`
In parallel, `i_(2) =E/(2 + r/2) = (2E)/(4 + r)`
Now, `i_(1) = i_(2) rArr (2E)/(2 + 2r) = (2E)/(4 + r)`
`rArr r = 2Omega`