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

Options

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

Answer

B

Explanation

(b) In series, `i_(1) = (2E)/(2 + 2r)`
In parallel, `I_(2) = (E)/(2 + (r )/(2)) = (2E)/(4 + r)`
Since `i_(1) = i_(2) implies (2E)/(4 + r) = (2E)/(2 + 2r) implies r = 2 Omega`