Question

A galvanometer together with an unknown resistance in series is connected across two identical batteries, each of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $1.5\text{V}$. When the batteries are connected in series, the galvanometer records a current of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $1\text{A}$, and when the batteries are in parallel the current is <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $0.6\text{A}$, what is the internal resistance of the battery?

• A. $1\Omega$
• B. $\frac{1}{2}\Omega$
• C. $\frac{1}{3}\Omega$
• D. $\frac{1}{4}\Omega$

Answer 1

The correct option is C $\frac{1}{3}\Omega$

The respective circuits are shown in figure (a) and (b)



If the internal resistance of the battery is $r$ then in the first case circuit current is given as,

$i=\frac{3}{R+2r}=1\text{A}$

$\Rightarrow R+2r=3$

$\Rightarrow R=(3-2r).....\left(1\right)$

In second case as batteries are in parallel equivalent EMF remains same at $1.5\text{V}$ and the two internal resistance are in parallel so their effective internal resistance is $r/2$ and the current in external resistance will be,

$i=\frac{E}{R+\frac{r}{2}}$

$\Rightarrow 0.6=\frac{E}{R+\frac{r}{2}}$

From equation $\left(1\right)$ substituting the value of $R$ gives,

$\Rightarrow 0.6=\frac{1.5}{(3-2r)+\frac{r}{2}}$

$\Rightarrow 3-1.5r=2.5$

$\therefore r=\frac{1}{3}\Omega$

Hence, option $\left(c\right)$ is correct answer.