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.
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.
CBSE DELHI - SET 1 2015
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$\text{I} = \frac{\text{E}}{\text{R} +\text{r}}$$1 = \frac{\text{E}}{4 + \text{r}}$
$ = >\text{E} = 4 +\text{r}$ ..... (i)
Also,
$0.5 =\frac{\text{E}}{9+ \text{r}}$
E = 4.5+ 0.5 r ..... (ii)
From equation (i) & (ii)
4 + r = 4.5 + 0.5
$\therefore\text{r} = 1\Omega$
Using this value of r, we get.
E = 5V.
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