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Q.

A wire of resistance 10 Ω 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 Ω as shown in the figure.  The currents in the two parts of the circle are

Question Image

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a

\frac{4}{{25}}A\,\,{\text{and}}\,{\text{ }}\frac{{12}}{{25}}A

b

\frac{5}{{26}}A\,\,{\text{and}}\,\frac{{15}}{{26}}A

c

\frac{6}{{23}}A\,\,{\text{and}}\,{\text{ }}\frac{{18}}{{23}}A

d

\frac{3}{{25}}A\,\,{\text{and}}\,{\text{ }}\frac{9}{{25}}A

answer is A.

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Detailed Solution

In the following figure Resistance of part PNQ;

Question Image
{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
Question Image
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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A wire of resistance 10 Ω 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 Ω as shown in the figure.  The currents in the two parts of the circle are