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Question

A wire of resistance 10Ω 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 3V and internal resistance 1Ω as shown in the figure. The currents in the two parts of the circle are
1024336_f1e326f9a9fa42b79ba01b54ce19fda4.PNG

A
623A and 1823A
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B
526A and 1526A
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C
425A and 1225A
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D
325A and 925A
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Solution

The correct option is A 623A and 1823A
when the wire is bent to form a circle arc PQ subtends 90 at the center. So, the resistance of the smaller part is 1/4th of 10Ω.
The resistance of the smaller arc is 2.5Ω and that of larger arc is 7.5Ω
They are connected in parallel to each other and is series to the internal resistance of the cell.
Total resistance=(2.5)(7.5)2+1=2.875
Now, total current in the circuit,=32.875
=1.04
The current in the two branches of the circle will split according to the inverse o ratio of resistance,i.e 1:3
i.e, 623 and 1823

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