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(A) The resistance of the wire is $R = 10\, \Omega$. The points $P$ and $Q$ divide the circle into two parts, one being a quadrant ($1/4$th of the cir

Vedclass · Accepted Answer

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

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(A) The resistance of the wire is $R = 10\, \Omega$. The points $P$ and $Q$ divide the circle into two parts, one being a quadrant ($1/4$th of the circumference) and the other being three-quarters ($3/4$th of the circumference).Resistance of the first part, $R_1 = \frac{1}{4} 	imes 10 = 2.5\, \Omega$.Resistance of the second part, $R_2 = \frac{3}{4} 	imes 10 = 7.5\, \Omega$.These two resistors are in parallel. Their equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2.5 	imes 7.5}{2.5 + 7.5} = \frac{18.75}{10} = 1.875\, \Omega = \frac{15}{8}\, \Omega$.The total resistance of the circuit including the internal resistance $r = 1\, \Omega$ is $R_{total} = R_{eq} + r = \frac{15}{8} + 1 = \frac{23}{8}\, \Omega$.The total current $I$ drawn from the battery is $I = \frac{V}{R_{total}} = \frac{3}{23/8} = \frac{24}{23}\, A$.Using the current divider rule, the current $i_1$ through $R_1$ is:$i_1 = I 	imes \frac{R_2}{R_1 + R_2} = \frac{24}{23} 	imes \frac{7.5}{10} = \frac{24}{23} 	imes 0.75 = \frac{18}{23}\, A$.The current $i_2$ through $R_2$ is:$i_2 = I - i_1 = \frac{24}{23} - \frac{18}{23} = \frac{6}{23}\, A$.

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