Question 72: Current Electricity

Question

Two cells of emfs 1 V and 2 V and internal resistance $2 \Omega$ and $1 \Omega$, respectively connected in parallel, gave a current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\frac{\alpha}{5} \mathrm{~A}$. The value of $\alpha$ is $\_\_\_\_$.

Accepted Answer

When two cells with emfs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ are connected in parallel, the equivalent $\operatorname{emf}\left(\varepsilon_{\mathrm{eq}}ight)$ and equivalent internal resistance $\left(\mathrm{r}_{\mathrm{eq}}ight)$ are :
$$ \varepsilon_{\mathrm{eq}}=\frac{\frac{\varepsilon_1}{\mathrm{r}_1}+\frac{\varepsilon_2}{\mathrm{r}_2}}{\frac{1}{\mathrm{r}_1}+\frac{1}{\mathrm{r}_2}}=\frac{\varepsilon_1 \mathrm{r}_2+\varepsilon_2 \mathrm{r}_1}{\mathrm{r}_1+\mathrm{r}_2} $$
$$ \mathrm{r}_{\mathrm{eq}}=\frac{\mathrm{r}_1 \mathrm{r}_2}{\mathrm{r}_1+\mathrm{r}_2} $$
Case 1 :
Cells are connected with same polarity
The voltages and internal resistances of two cells are $\varepsilon_1=1 \mathrm{~V}, \mathrm{r}_1=2 \Omega$ and $\varepsilon_2=2 \mathrm{~V}, \mathrm{r}_2=1 \Omega$.
$$ \varepsilon_{\mathrm{eq}}=\frac{(1 	imes 1)+(2 	imes 2)}{2+1}=\frac{1+4}{3}=\frac{5}{3} \mathrm{~V} $$
The equivalent internal resistance is,
$$ \mathrm{r}_{\mathrm{eq}}=\frac{2 	imes 1}{2+1}=\frac{2}{3} \Omega $$
The current I through external resistance R is :
$$ I=\frac{\varepsilon_{\mathrm{eq}}}{R+r_{\mathrm{eq}}} $$
$$\Rightarrow $$ $$ 1=\frac{5 / 3}{\mathrm{R}+\frac{2}{3}} \Rightarrow \mathrm{R}+\frac{2}{3}=\frac{5}{3} \Rightarrow \mathrm{R}=1 \Omega $$
Case 2
: Polarity of One Cell Reversed
If we reverse the 1 V cell, its emf becomes -1 V .
The equivalent emf of combination is,
$$ \varepsilon_{\mathrm{eq}}^{\prime}=\frac{(-1 	imes 1)+(2 	imes 2)}{2+1}=\frac{-1+4}{3}=\frac{3}{3}=1 \mathrm{~V} $$
The equivalent internal resistance $\mathrm{r}_{\mathrm{eq}}$ remains $\frac{2}{3} \Omega$.
So, the new current $\mathrm{I}^{\prime}$ across the external resistance is:
$$ I^{\prime}=\frac{\varepsilon_{\mathrm{eq}}^{\prime}}{R+r_{\mathrm{eq}}}=\frac{1}{1+\frac{2}{3}}=\frac{1}{5 / 3}=\frac{3}{5} \mathrm{~A} $$
It is given that $\mathrm{I}^{\prime}=\frac{\alpha}{5}$ :
$$ \frac{3}{5}=\frac{\alpha}{5} \Rightarrow \alpha=3 $$
Therefore, the value of $\alpha$ is 3 .

Current Electricity

Integer / Numeric Answer Type