# A certain length of a uniform wire of resistance 12 ohm is bent into a circle and two points, a quarter of circumference apart ,are connected to a battery of EMF 4 volt and internal resistance 1 ohm find the current in the different parts of the circuit with diagram

Let the length of the uniform wire be L.

The resistance per unit length of the wire will be 12/L

It is bent in the form of a circle of radius,r such that

$2\pi r=L\phantom{\rule{0ex}{0ex}}r=\frac{L}{2\pi}$

A battery is connected across two point separated by a quarter of the circle. So, across these two points there will be two parallel resistances, one due to 1/4th length of the wire, R1 and the other due to 3/4th length of the wire, R2

${R}_{1}=\frac{12}{2\pi r}\times \frac{2\pi r}{4}=3\Omega \phantom{\rule{0ex}{0ex}}{R}_{2}=\frac{12}{2\pi r}\times \frac{3}{4}\times 2\pi r=9\Omega \phantom{\rule{0ex}{0ex}}So,{R}_{1}\parallel {R}_{2}=R=\frac{9}{3+1}=\frac{9}{4}\Omega $

Net current through the wire:

$i=\frac{E}{R+r}=\frac{4}{{\displaystyle \frac{9}{4}}+1}=\frac{16}{13}A=1.23A\phantom{\rule{0ex}{0ex}}Currentthrough{R}_{1}i.e.,quarterlengthofthewire\phantom{\rule{0ex}{0ex}}{i}_{1}=\frac{9}{12}\times \frac{16}{13}A=\frac{12}{13}=0.92A\phantom{\rule{0ex}{0ex}}Currentthrough{R}_{2}i.e.,quarterlengthofthewire\phantom{\rule{0ex}{0ex}}{i}_{2}=\frac{3}{12}\times 1.23=\frac{4}{13}=0.31A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Regards

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