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
A
\(\frac{V_2}{I}\)
B
\(\frac{V_1}{I}\)
C
\(\frac{(V_1 + V_2)}{I}\)
D
\(\frac{(V_1 - V_2)}{I}\)
E
\(\frac{(V_2 - V_1)}{I}\)
Verified Correct Answer: D.
Physical Principles & Numerical Solution
On open circuit, voltage = V\(_1\), On a closed circuit voltage drops due to internal resistance V\(_2\) = I r
But according to Ohm's law, voltage drop( V\(_d\) = I r
V\(_d\) = V\(_1\) - V\(_2\) = Ir
r = \(\frac{(V_1 - V_2)}{I}\)
Source: Myschool Past Questions Archive (Record #29130)