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(C) To find the equivalent resistance,we simplify the circuit step-by-step:$1$. On the left side,the $10\, \Omega$ and $15\, \Omega$ resistors are in

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$5\, \Omega$

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(C) To find the equivalent resistance,we simplify the circuit step-by-step:$1$. On the left side,the $10\, \Omega$ and $15\, \Omega$ resistors are in parallel. Their equivalent resistance $R_1$ is given by $\frac{1}{R_1} = \frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6}$,so $R_1 = 6\, \Omega$.$2$. On the right side,the $8\, \Omega$ and $8\, \Omega$ resistors are in parallel. Their equivalent resistance $R_2$ is $\frac{8 	imes 8}{8+8} = \frac{64}{16} = 4\, \Omega$.$3$. Now,the $6\, \Omega$ resistor is in series with $R_2$ $(4\, \Omega)$,giving a total resistance $R_3 = 6 + 4 = 10\, \Omega$ on the right branch.$4$. The $4\, \Omega$ resistor on the bottom is in series with $R_1$ $(6\, \Omega)$,giving a total resistance $R_4 = 4 + 6 = 10\, \Omega$ on the left branch.$5$. Finally,the two branches of $10\, \Omega$ each are in parallel across the $24\, V$ source. The equivalent resistance $R_{eq}$ is $\frac{10 	imes 10}{10 + 10} = \frac{100}{20} = 5\, \Omega$.

Find the equivalent resistance across the terminals of the source of $e.m.f.$ $24\, V$ for the circuit shown in the figure.

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