Question

A wire of resistance $10\Omega$ is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant and are connected to a battery of $3V$ and internal resistance $1\Omega$ as shown in the figure. The currents in the two parts of the circle are

• A. $\frac{6}{23}A$ and $\frac{18}{23}A$
• B. $\frac{5}{26}A$ and $\frac{15}{26}A$
• C. $\frac{4}{25}A$ and $\frac{12}{25}A$
• D. $\frac{3}{25}A$ and $\frac{9}{25}A$

Answer 1

The correct option is A $\frac{6}{23}A$ and $\frac{18}{23}A$

when the wire is bent to form a circle arc $PQ$ subtends $90$ at the center. So, the resistance of the smaller part is $1/4th$ of $10\Omega$.

The resistance of the smaller arc is $2.5\Omega$ and that of larger arc is $7.5\Omega$

They are connected in parallel to each other and is series to the internal resistance of the cell.

Total resistance=$\frac{\left(2.5\right)\left(7.5\right)}{2}+1=2.875$

Now, total current in the circuit,$=\frac{3}{2.875}$

$=1.04$

The current in the two branches of the circle will split according to the inverse o ratio of resistance,i.e $1:3$

i.e, $\frac{6}{23}$ and $\frac{18}{23}$