Question

A galvanometer, together with an unknown resistance in series, is connected across two identical batteries of each $1.5V$. When the batteries are connected in series, the galvanometer records a current of $1A,$ and when the batteries are connected in parallel, the current is $0.6A$ , then the internal resistance is $1/x^{\prime }\Omega$. What is the value of $x$'?

Answer 1

Let $R$ be the combined resistance of galvanometer and an unknown resistance in series and $r$ be the internal resistance of each battery. When the batteries, each of emf $E$ are connected in series, the net emf $=$ $2E$ and net internal resistance $=$ $2r$.
Current $i_1=\frac{2E}{R+2r}or1.0=\frac{2\times 1.5}{R+2r}$
or $R+2r=3$ (i)
When the batteries are connected in parallel, the emf remains $E$ and net internal resistance become $r/2$. Therefore, current is:
$i_2=\frac{E}{R+\frac{r}{2}}=\frac{2E}{2R+r}$
or $2R+r=\frac{2E}{i_2}=\frac{2\times 1.5}{0.6}=5.0$ (ii)
Solving Equations (i) and (ii), we get $r=\frac{1}{3}\Omega$

From question, Value of $x=3$