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(A) The total resistance of the wire is $R = 10 \, \Omega$. The points $P$ and $Q$ divide the circle into two parts: a quadrant (arc $PNQ$) and the re

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The total resistance of the wire is $R = 10 \, \Omega$. The points $P$ and $Q$ divide the circle into two parts: a quadrant (arc $PNQ$) and the remaining three-fourths (arc $PMQ$).Resistance of the quadrant part $PNQ$ is $R_1 = \frac{1}{4} 	imes 10 = 2.5 \, \Omega$.Resistance of the remaining part $PMQ$ is $R_2 = \frac{3}{4} 	imes 10 = 7.5 \, \Omega$.These two parts are connected 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 main current $i$ 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:$i_1 = i 	imes \left( \frac{R_2}{R_1 + R_2} ight) = \frac{24}{23} 	imes \left( \frac{7.5}{2.5 + 7.5} ight) = \frac{24}{23} 	imes 0.75 = \frac{24}{23} 	imes \frac{3}{4} = \frac{18}{23} \, A$.The current $i_2$ through $R_2$ is:$i_2 = i - i_1 = \frac{24}{23} - \frac{18}{23} = \frac{6}{23} \, A$.Thus, the currents are $\frac{6}{23} \, A$ and $\frac{18}{23} \, A$.

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