Question

The equivalent resistance of the circuit across $\text{AB}$ is given by

• A. $4\Omega$
• B. $13\Omega$
• C. $4\Omega$ or $13\Omega$
• D. $4\Omega$ or zero

Answer 1

The correct option is C $4\Omega$ or $13\Omega$


$\text{Case 1}:V_A>V_B$

In this case, the diodes are forward biased, then the resistance of each of the diodes is taken to be zero. Thus, the circuit reduces to


As $\frac{2}{4}=\frac{4}{8}$, the circuit satisfies the Wheatstone bridge condition, and we can remove the $5\Omega$ resistor.

Then the resultant circuit can be shown as


$\therefore R_{eq}=\frac{6\times 12}{6+12}=4\Omega$

$\text{Case 2}:V_A<V_B$

For this case diodes become reverse biased, then the section having the diodes can be removed as the diodes possess infinite resistance.


In this case, $4\Omega ,5\Omega$ and $4\Omega$ are in series.

$\therefore R_{eq}=4+5+4=13\Omega$

Hence, option $\left(\text{C}\right)$ is correct.