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
Medium
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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 ${i_2} = i - {i_1} = \frac{{24}}{{23}} - \frac{{18}}{{23}} = \frac{6}{{23}}A$.
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