Q. A cell of emf E is connected across a resistance R. the potential difference between the terminals of the cell is found to be V volt. Then the internal resistance of the cell must be
A) (E-V)
B) $\frac{(E - V)}{V}R$
C) $\frac{2(E - V)R}{E}$
D) $\frac{2(E - V)V}{R}$
Explanation
(b) Let the potential difference across battery is V and internal resistance of the cell is r, then
$E = V + ir\ \ \ \ \ldots.(i)$
$V = iR\ \ \ \ \ \ldots\ldots\ldots(ii)$
Now, from Eqs. (i) and (ii) we have
$E = iR + ir = i(R + r)$ …..(iii)
Now, dividing Eq. (iii) by Eq. (ii), we get
$\frac{E}{V} = \frac{R + r}{R} = 1 + \frac{r}{R}\ $
$\frac{E}{V} - 1 = \frac{r}{R}$
$or\ \ \ \left( \frac{E - V}{V} \right)R = r$
Hence, internal resistance
$r = \left( \frac{E - V}{V} \right)R\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.