Question

A galvanometer connected with an unknown resistor and two identical cells in series each of emf $2\, V$ shows a current of $1\, A$. If the cells are connected in parallel, it shows $0.8\, A$. Then the internal resistance of the cell is

• A. $1\, \Omega$
• B. $0.5\, \Omega$
• C. $0.25\, \Omega$
• D. $0.33\, \Omega$

Answer 1

Answer: (A)

Here, emf of each cell, $\varepsilon=2 V$ Let $R$ be resistance of unknown resistor and $r$ be internal resistance of each cell.

In circuit (a), When the cells are connected in series, Current in the circuit,
$I=\frac{2 \varepsilon}{r+ r+ R}=\frac{2 \times 2}{2 r +R}$
$1=\frac{4}{2 r+ R}$ or $2 r+ R=4$...(i)

In circuit (b), When the cells are connected in parallel, Current in the circuit,
$I=\frac{\varepsilon}{\frac{r \times r}{r+ r}+R}=\frac{2}{\frac{r}{2}+R}$
$0.8=\frac{2}{\frac{r}{2}+ R}$
$\frac{r}{2}+R=\frac{2}{0.8}=\frac{5}{2}$...(ii)
Subtract (ii) from (i), we get
$\frac{3}{2} r=\frac{3}{2}$ or $r=1\, \Omega$