- Two cells of emf $E_1$ and $E_2$ have their internal resistances $r_1$ and $r_2$, respectively. Deduce an expression for the equivalent emf and internal resistance of their parallel combination when connected across an external resistance $R.$ Assume that the two cells are supporting each other.
- In case the two cells are identical, each of emf $E = 5V$ and internal resistance $\text{r}=2\Omega,$ calculate the voltage across the external resistance $\text{R}=10\Omega.$
CBSE 55-1-3 PAPER SET 2020
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$\text{V}=\text{V}_\text{A}-\text{V}_\text{B}=\text{E}_1-\text{I}_1\text{r}_1\ ...(1)$
$\text{V}=\text{V}_\text{A}-\text{V}_\text{B}=\text{E}_2-\text{I}_2\text{r}_2\ ...(2)$
$\Rightarrow\text{I}_1=\frac{\text{E}_1}{\text{r}_1}-\frac{\text{V}}{\text{r}_1}\ ...(3)$ (from (1))
$\text{I}_2=\frac{\text{E}_2}{\text{r}_2}-\frac{\text{V}}{\text{r}_2}\ ...(4)$ (from (2))
For Equivalent cell $\text{I}=\frac{\text{E}}{\text{r}}-\frac{\text{V}}{\text{r}}\ ...(5)$
$\because\text{I}=\text{I}_1+\text{I}_2$
$\therefore\frac{\text{E}}{\text{r}}\frac{\text{V}}{\text{r}}=\Big(\frac{\text{E}_1}{\text{r}_1}-\frac{\text{V}}{\text{r}_1}\Big)+\Big(\frac{\text{E}_2}{\text{r}_2}-\frac{\text{V}}{\text{r}_2}\Big)$
$=\Big(\frac{\text{E}_1}{\text{r}_1}+\frac{\text{E}_2}{\text{r}_2}\Big)-\text{V}\Big(\frac{1}{\text{r}_1}+\frac{1}{\text{r}_2}\Big)$
Comparing we get $\frac{1}{\text{r}}=\frac{1}{\text{r}_1}+\frac{1}{\text{r}_2}$
$\therefore$ Equivalent internal resistance is, $\text{r}=\frac{\text{r}_1\text{r}_2}{\text{r}_1+\text{r}_2}$
Also, $\frac{\text{E}}{\text{r}}=\frac{\text{E}_1}{\text{r}_1}+\frac{\text{E}_2}{\text{r}_2}=\frac{\text{E}_1\text{r}_2+\text{E}_2\text{r}_1}{\text{r}_1\text{r}_2}$
$\therefore$ Equivalent emf is, $\text{E}=\frac{\text{E}_1\text{r}_2+\text{E}_2\text{r}_1}{\text{r}_1+\text{r}_2}$
- $\text{E}=\frac{5\times2+5\times2}{2+2}=5\text{V}$
$\text{I}=\frac{\text{E}}{\text{R}+\text{r}}=\frac{5}{10+1}=\frac{5}{11}\text{A}$
$\therefore$ Voltage across $\text{R}\Rightarrow\text{V}$
$=\text{IR}=\frac{5}{11}\times10=\frac{50}{11}\text{V}=4.54\text{V}$
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