A cell has an open circuit voltage V1 and a closed circuit voltage V2 when a current is flowing through it. Therefore, the internal resistance of the cell is
\(\frac{V_2}{I}\)
\(\frac{V_1}{I}\)
\(\frac{(V_1 + V_2)}{I}\)
\(\frac{(V_1 - V_2)}{I}\)
\(\frac{(V_2 - V_1)}{I}\)
Explanation
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Discussions (13)

The correct answer is D because
From the relationship between EMF, Potential difference and the lost voltage we have that
E = V+v -------(1)
Where E is the open circuit voltage when it is not driving any current to the external circuit
Where V is the closed circuit voltage the cell is now supplying current to the external circuit now.
E - V = v ----------(2)
Where v = Ir ----------(3)
Substituting eqn (3) to (2)
r = E-V/I

My school the answer should be D cuz when its an open circuit its an electromotive force where the circuit does no work. So it should be v1-v2/1

Someone should please explain
I thought that in an open circuit current cannot flow through it





