Question

The Equivalent Resistance between ${\text{A}}$ and ${\text{B}}$ is

A) $36\Omega$ if ${{\text{V}}_{\text{A}}}{\text{ > }}{{\text{V}}_{\text{B}}}$
B) $18\Omega$ if ${{\text{V}}_{\text{A}}}{\text{ > }}{{\text{V}}_{\text{B}}}$
C) $0$ if ${{\text{V}}_{\text{A}}}{\text{ > }}{{\text{V}}_{\text{B}}}$
D) $54\Omega$ if ${{\text{V}}_{\text{A}}}{\text{ > }}{{\text{V}}_{\text{B}}}$

Answer 1

In Order to Solve this Question, we have to draw an Equivalent circuit, in which we neglect the diode, and connect them directly to the resistance and simply calculate the resistance with taking two cases.
${\text{I}}$ case ${{\text{V}}_{\text{A}}}{\text{ > }}{{\text{V}}_{\text{B}}}$
${\text{II}}$ case ${{\text{V}}_{\text{A}}}{\text{ < }}{{\text{V}}_{\text{B}}}$

Formula Used:
${{\text{R}}_{{\text{equivalent}}}}{\text{ = }}{{\text{R}}_{\text{1}}}{\text{ + }}{{\text{R}}_{\text{2}}}$ [in series]

Complete step by step Solution:
Firstly, we calculate the Resistance by considering Potential of ${\text{A}}$ is greater than that of ${\text{B}}$ i.e. ${{\text{V}}_{\text{A}}}{\text{ > }}{{\text{V}}_{\text{B}}}$.

Here the Diodes are in forward Bias & when they are connected as a forward Bias. The circuit behaves like a short- circuit, and current follows the path.
In which resistance is less.

In Fig $\left( 2 \right)$ the current flows through the path AP & then to PQ & then QB, (Because it follows the path where Resistance is less.)
So, when ${{\text{V}}_{\text{A}}}{\text{ > }}{{\text{V}}_{\text{B}}}$
${\text{ }}{{\text{R}}_{{\text{eq}}}}$ is equal to $0$. Because there is no resistance in the path.\Now we will consider the ${{\text{2}}^{{\text{nd}}}}$ case when ${{\text{V}}_{\text{A}}}{\text{ < }}{{\text{V}}_{\text{B}}}$.
In this case ${{\text{V}}_{\text{A}}}$ is at negative Potential and ${{\text{V}}_{\text{B}}}$ is at positive Potential.

Here the diode X and Y are in Reverse Bias. Only, Diode Z diode is at forward Bias. Hence whenever there is a Reverse Bias current do not flow through it. So the current will flow through Path ${\text{AD}}$ where Resistance $18\Omega$ is connected.

So, in Fig. $3,$ the Equivalent Resistance is given by
${{\text{R}}_{{\text{eq}}}}{\text{ = }}{{\text{R}}_{\text{1}}}{\text{ + }}{{{R}}_{\text{2}}} \\\ {{ = 18\Omega + 36\Omega }} \\\ {{ = 54\Omega [In - series]}} \\\$

Hence, Option (C) is correct.

Note: For the case where A is at a positive potential and B is at negative potential, the entire current will flow through the Diode, this is why as the resistors are not included in the circuit we can assume them to be disconnected. However, if the point A is at negative potential and the point B is at positive potential the Diode is now reverse biased. therefore it behaves as a conducting wire and the resistors become like series combinations.