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(C) From the given figure,we can observe that the two resistors of $4 \ \Omega$ and $8 \ \Omega$ in the upper branch are in series. Their equivalent r

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$16$

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(C) From the given figure,we can observe that the two resistors of $4 \ \Omega$ and $8 \ \Omega$ in the upper branch are in series. Their equivalent resistance is $R_1 = 4 \ \Omega + 8 \ \Omega = 12 \ \Omega$.This $12 \ \Omega$ resistor is in parallel with the $6 \ \Omega$ resistor connected between the same two nodes. The equivalent resistance $R_p$ of this parallel combination is given by $\frac{1}{R_p} = \frac{1}{12} + \frac{1}{6} = \frac{1+2}{12} = \frac{3}{12} = \frac{1}{4}$,so $R_p = 4 \ \Omega$.Now,the circuit simplifies to a series combination of the lower $4 \ \Omega$ resistor,the equivalent $4 \ \Omega$ resistor,and the lower $8 \ \Omega$ resistor.Therefore,the total equivalent resistance between $A$ and $B$ is $R_{AB} = 4 \ \Omega + 4 \ \Omega + 8 \ \Omega = 16 \ \Omega$.

The equivalent resistance between $A$ and $B$ for the mesh shown in the figure is ......... $\Omega$.

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