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(C) To find the equivalent resistance between points $A$ and $B$,we simplify the circuit step-by-step.$1$. The two $5 \, \Omega$ resistors in the top-

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$5$

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(C) To find the equivalent resistance between points $A$ and $B$,we simplify the circuit step-by-step.$1$. The two $5 \, \Omega$ resistors in the top-left branch are in series,giving $5 \, \Omega + 5 \, \Omega = 10 \, \Omega$.$2$. This $10 \, \Omega$ equivalent resistance is in parallel with the $10 \, \Omega$ resistor connected to point $A$. Their equivalent resistance is $\frac{10 	imes 10}{10 + 10} = 5 \, \Omega$.$3$. Now,this $5 \, \Omega$ equivalent resistance is in series with the next $5 \, \Omega$ resistor,giving $5 \, \Omega + 5 \, \Omega = 10 \, \Omega$.$4$. This $10 \, \Omega$ is in parallel with the next $10 \, \Omega$ resistor,giving $\frac{10 	imes 10}{10 + 10} = 5 \, \Omega$.$5$. Finally,this $5 \, \Omega$ is in series with the last $5 \, \Omega$ resistor,giving $5 \, \Omega + 5 \, \Omega = 10 \, \Omega$.$6$. This $10 \, \Omega$ is in parallel with the $10 \, \Omega$ resistor connected directly between $A$ and $B$. The final equivalent resistance is $R_{AB} = \frac{10 	imes 10}{10 + 10} = 5 \, \Omega$.

The equivalent resistance between points $A$ and $B$ in the given network is ............ $\Omega$.

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