Q. 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 $3\ V$ 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 $\frac{9}{25}A$
Explanation
(a) In the following figure
Resistance of part $PNQ;$
$R_{1} = \frac{10}{4} = 2.5\Omega$ and
Resistance of part $PMQ;$
$R_{2} = \frac{3}{4} \times 10 = 7.5\Omega$
$R_{eq} = \frac{R_{1}R_{2}}{R_{1} + R_{2}} = \frac{2.5 \times 7.5}{(2.5 + 7.5)} = \frac{15}{8}\Omega$
Main Current $i = \frac{3}{\frac{15}{8} + 1} = \frac{24}{23}A$
So, $i_{1} = i \times \left( \frac{R_{2}}{R_{1} + R_{2}} \right) = \frac{24}{23} \times \left( \frac{7.5}{2.5 + 7.5} \right) = \frac{18}{23}A$
and $i_{2} = i - i_{1} = \frac{24}{23} - \frac{18}{23} = \frac{6}{23}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.