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 $1A$ and when the batteries are in parallel, the current is $0.6A$. The internal resistance of the batteries is

• A. $2/3\Omega$
• B. $1\Omega$
• C. $4/3\Omega$
• D. $1/3\Omega$

Answer 1

The correct option is D $1/3\Omega$

When batteries are in series , the equivalent resistance of the circuit is $R_S=R+(r+r)=R+2r$ and
current, $I_S=\frac{E+E}{R_S}=1$
or $\frac{2E}{R+2r}=1$
or $R+2r=2\times 1.5=3\Rightarrow R=3-2r...\left(1\right)$
When batteries are in parallel , the equivalent resistance of the circuit is $R_S=R+\left(r||r\right)=R+r/2$ and
current, $I_P=\frac{E}{R_P}=0.6$
or $\frac{E}{R+r/2}=0.6$
or $R+r/2=E/0.6=1.5/0.6=5/2$
using (1), $3-2r+r/2=5/2$ or $3r/2=1/2$
$\Rightarrow r=1/3\Omega$