Question

A cell of emf 'E' and internal resistance 'r' is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current I.
It is found that When $R=4\Omega$, the current is 1 A and when R is increased to $9\Omega$, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.

Answer 1

current in the circuit is given by

$I=\frac{\epsilon }{R+r}$

Terminal voltage $V=IR=V_A-V_{B}V=\frac{\epsilon I}{R+r}=\epsilon -IrI=\frac{\epsilon }{4+r}⟹r+4=\epsilon ⟶\left(1\right)0.5=\frac{\epsilon }{9+r}0.5r+4.5=\epsilon ⟶\left(2\right)$

On solving (1) and (2), we have $\epsilon =5V$ EMF

$R=5-4=1\Omega$ Internal resistance