Question

A wire of resistance 10 ohm is bent to form a circle. P & Q are points on the circumference of the circle dividing it onto a quadrant and are connected to a battery of 3 volt an internal resistance of 1 ohm as shown on the figure. The current in the two parts of the circle are

Answer 1

Given,

$Restance=10ohm$

$Battery=3v$

Internal resistance=$1ohm$

So, when the wire is bent to form a circle are PQ substend ${90}^0$ at the center of the resistance of smaller part is $\frac{1}{4}of10\Omega$

And the bigger part is $\frac{3}{4}of10\Omega$

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

They are connected in the parallel each other and in series to be internal ressistance,

Total resistance in the cercuit is equal to the sum of the resistance in parallel and the internal resistance of the cell.

Total resistance is $\frac{\left(2.5\right)\left(7.5\right)}{10}+1$

$=2.875$

Now total current in the circuit is:

$\frac{3}{2.875}=\frac{24}{23}$

So, the current in the two branches of the circle will split according to the inverse of ratio of resistance

$1:3$ i;e $\frac{6}{23}\$\frac{18}{23}$