Question

In the circuit shown in the figure, a meter bridge is in its balance state. The meter bridge wire has a resistance of $0.1$$\Omega$/$\text{cm}$. Calculate the value of unknown resistance $X$ and the current drawn from the battery of negligible internal resistance.

Answer 1

Since, this is a balanced meter bridge, so

$\frac{X}{40}=\frac{3}{60}$

$\therefore X=2\Omega$

Since, meter bridge is balanced so we can remove the galvanometer.
Now the simplified circuit will be :

$X+3=2+3=5\Omega$

$R_{AB}$ $=$ $0.1$ $\Omega$/$\text{cm}$ $\times$ $100\text{cm}=10\Omega$

$5\Omega$ and $10\Omega$ are in parallel, so equivalent resistance will be

$R_{eq}=\frac{5\times 10}{5+10}=\frac{10}{3}\Omega$

So, current drawn from battery

$i=V/R_{eq}=\frac{6}{\left(10/3\right)}=1.8\text{A}$

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