Question

Two batteries of emf ${\epsilon }_1$ and ${\epsilon }_2$ having internal resistance $r_1$ and $r_2$ respectively are connected in series to an external resistance R. Both the batteries are getting discharged. The given described combination of these two batteries has to produce a weaker current than when any one of the batteries is connected to the same resistor. For this requirement to be fulfilled.

• A. $\frac{{\epsilon }_2}{{\epsilon }_1}$ must not lie between $\frac{-r_2}{r_1+R}$ and $\frac{r_1}{r_2+R}$
• B. $\frac{{\epsilon }_2}{{\epsilon }_1}$ must not lie between $\frac{r_2}{r_1+R}$ and $\frac{r_2+R}{r_1}$
• C. $\frac{{\epsilon }_2}{{\epsilon }_1}$ must lie between $\frac{r_2}{r_1+R}$ and $\frac{r_1}{r_2+R}$
• D. $\frac{{\epsilon }_2}{{\epsilon }_1}$ must lie between $\frac{r_2}{r_1+R}$ and $\frac{r_2+R}{r_1}$

Answer 1

The correct option is B $\frac{{\epsilon }_2}{{\epsilon }_1}$ must not lie between $\frac{r_2}{r_1+R}$ and $\frac{r_2+R}{r_1}$

Ref. image 1

Given two cells of emf ${\epsilon }_1$ and ${\epsilon }_2$ are joined in series such that current, $I=\frac{E_1+E_2}{r_1+r_2+R}$____(1)

When any one of battery is connected with the resistor

Ref. image 2

then the current can be given as, $I_1=\frac{{\epsilon }_1}{r_1+R}$ ___(2)

When other battery is connected

Ref. image 3

then the current can be given as, $I_2=\frac{{\epsilon }_2}{r_2+R}$ ___(3)

According to the questions as per the given condition, the current $I$ should be less than any of the individual current, $I1$ or $I2$:

$I<I_1$____(4)

$I<I_2$____(5)

From eq. $\left(1\right)\left(2\right)and\left(4\right)$

$\frac{{\epsilon }_1+{\epsilon }_2}{R+r_1+r_2}<\frac{{\epsilon }_1}{r_1+R}$

On rearranging, the above eq. can be written as:

$\frac{{\epsilon }_1+{\epsilon }_2}{{\epsilon }_1}<\frac{R+r_1+r_2}{r_1+R}$

$\Rightarrow \frac{{\epsilon }_2}{{\epsilon }_1}+1<\frac{R+r_1}{r_1+R}+\frac{r_2}{r_1+R}$

$\Rightarrow \frac{{\epsilon }_2}{{\epsilon }_1}+1<1+\frac{r_2}{r_1+R}$

$\Rightarrow \frac{{\epsilon }_2}{{\epsilon }_1}<\frac{r_2}{r_1+R}$____(6)

From eq. $\left(1\right)\left(3\right)and\left(5\right)$:

$\frac{{\epsilon }_1+{\epsilon }_2}{R+r_1+r_2}<\frac{{\epsilon }_2}{r_2+R}$

On rearranging, the above eq. can be written as:

$\frac{{\epsilon }_1+{\epsilon }_2}{{\epsilon }_2}<\frac{R+r_1+r_2}{r_2+R}$

$\Rightarrow \frac{{\epsilon }_1}{{\epsilon }_2}+1<\frac{R+r_2}{r_2+R}+\frac{r_1}{r_2+R}$

$\Rightarrow \frac{{\epsilon }_1}{{\epsilon }_2}+1<1+\frac{r_1}{r_2+R}$

$\Rightarrow \frac{{\epsilon }_1}{{\epsilon }_2}<\frac{r_1}{r_2+R}$

By taking the reciprocal and using inequality, we can write:

$\Rightarrow \frac{{\epsilon }_2}{{\epsilon }_1}<\frac{r_2+R}{r_1}$____(7)

From equation $\left(6\right)and\left(7\right)$, we can see that

$\frac{{\epsilon }_2}{{\epsilon }_1}<\frac{r_2}{r_1+R}$ $and$ $\frac{{\epsilon }_2}{{\epsilon }_1}<\frac{r_2+R}{r_1}$

Which concludes $\frac{{\epsilon }_2}{{\epsilon }_1}$ must not lie between $\frac{r_2}{r_1+R}$ and $\frac{r_2+R}{r_1}$
Hence, the correct answer is option $\left(B\right)$