Question

In the circuit shown below, $V_A$ and $V_B$ are the potentials at A and B, R is the equivalent resistance between A and B, $S_1$ and $S_2$ are switches, and the diodes are ideal

• A. if $V_A>V_B,S_1$ is open and $S_2$ is closed then $R=8\Omega$
• B. if $V_A>V_B,S_1$ is closed and $S_2$ is open then $R=12.5\Omega$
• C. if $V_A>V_B,S_1$ is open and $S_2$ is closed then $R=12.5\Omega$
• D. if $V_A>V_B,S_1$ is closed and $S_2$ is open then $R=8\Omega$

Answer 1

Answer: (A), (B)

The correct options are
A if $V_A>V_B,S_1$ is open and $S_2$ is closed then $R=8\Omega$
B if $V_A>V_B,S_1$ is closed and $S_2$ is open then $R=12.5\Omega$

Given $V_A>V_B$ ,so current flows from A to B.

From the below figure when

$S_1$ is open and $S_2$ is closed:-

The diode at $S_2$ passes current only from D to C or else zero current.

If current passes through diode then voltage at D and C will be equal.

Let us assume that voltages are equal at D and C. Now we can short the $S_2$ part. Now in the equivalent circuit as 20$\Omega >5\Omega$ current flows from D to C.

Thus it is satisfying the diode property.

The equivalent circuit of the above case will be our circuit. So the equivalent resistance

$=2\times \frac{1}{(\frac{1}{20}+\frac{1}{5})}\Omega$

$=8\Omega$

$S_2$ is open and $S_1$ is closed:-

The diode at $S_1$ passes current only from C to D or else zero current.

If current passes through diode then voltage at D and C will be equal.

Let us assume that voltages are equal at D and C. Now we can short the $S_1$ part. Now in the equivalent circuit as 20$\Omega >5\Omega$ current has to flow from D to C. But diode does not allow current from D to C. So our assumption is wrong.

We have to treat $S_1$ as open. The equivalent resistance is

$=\frac{20+5}{2}\Omega$

$=12.5\Omega$