Question

A wire of resistance $16\Omega$ is bent to form a closed circle. Two points on its circumference, situated quarter of the circumference apart, are connected to a cell of emf 2 V and internal resistance $1\Omega$. Calculate the current in each part of the circuit.

Answer 1

$R_{eq}={(\frac{1}{4}+\frac{1}{12})}^{-1}+1=\frac{\text{/}4\times 12}{\text{/}{16}_4}+1=4\Omega I=\frac{2}{4}=0.5A{I}_1=\left(\frac{12}{16}\right)0.5=0.375A{I}_2=\left(\frac{4}{16}\right)0.5=0.125A$