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(D) In the given circuit,the two $4\,\Omega$ resistors are connected in parallel with a short-circuiting wire. This means the current will bypass thes

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$10$

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(D) In the given circuit,the two $4\,\Omega$ resistors are connected in parallel with a short-circuiting wire. This means the current will bypass these resistors,effectively removing them from the circuit.After removing the shorted resistors,the circuit simplifies to a $3\,\Omega$ resistor in series with a parallel combination of a $2\,\Omega$ resistor and another $2\,\Omega$ resistor,which is then in series with a $6\,\Omega$ resistor.The equivalent resistance of the two $2\,\Omega$ resistors in parallel is:$R_p = \frac{2 	imes 2}{2 + 2} = \frac{4}{4} = 1\,\Omega$Now,the total equivalent resistance $R_{eq}$ between terminals $A$ and $B$ is the sum of the series components:$R_{eq} = 3\,\Omega + R_p + 6\,\Omega$$R_{eq} = 3\,\Omega + 1\,\Omega + 6\,\Omega = 10\,\Omega$

In the given circuit,the equivalent resistance between the terminals $A$ and $B$ is $........\Omega$.

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