question_answer

  A wire of resistance 10 W 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 W as shown in the figure.  The currents in the two parts of the circle are                              [Roorkee 1999]

A)            \[\frac{6}{23}A\,\,\text{and}\,\text{ }\frac{18}{23}A\]

B)            \[\frac{5}{26}A\,\,\text{and}\,\frac{15}{26}A\]

C)            \[\frac{4}{25}A\,\,\text{and}\,\text{ }\frac{12}{25}A\]

D)            \[\frac{3}{25}A\,\,\text{and}\,\text{ }\frac{9}{25}A\]

Correct Answer: A

Solution :

                   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, i1= \[i\times \left( \frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)=\frac{24}{23}\times \left( \frac{7.5}{2.5+7.5} \right)=\frac{18}{23}A\] and \[{{i}_{2}}=i-{{i}_{1}}=\frac{24}{23}-\frac{18}{23}=\frac{6}{23}A\].