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# In the circuit shown below, ${V_A}$and ${V_B}$ are the potential 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 $R = 8\Omega$

Last updated date: 19th Jun 2024
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Hint:An electric circuit and each component of the circuit are known as an element. The given circuit is connected from the resistance values $20\Omega$ and$5\Omega$.
The ideal diodes are connected from the ${S_{1{\text{ }}}}{\text{and }}{S_2}$ switches.

It is 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 else zero current.
-If the current passes through the diode the voltage will be equal to D and C.

Let us assume that the D and C voltages are equal. Now we can short the ${S_2}$part.
Now in the equivalent circuit as$20\Omega > 5\Omega$current flows from D to C.
So it is satisfying the diode property.
Our circuit equivalent circuit will be the above circuit.
So the equivalent resistance $= 2 \times \dfrac{1}{{\left( {\dfrac{1}{{20}} + \dfrac{1}{5}} \right)}}\Omega$
Here we have to take an LCM on the denominator term we get
$= 2 \times \dfrac{1}{{\left( {\dfrac{{1 + 4}}{{20}}} \right)}}\Omega$
Let us add the term we get
$= 2 \times \dfrac{1}{{\left( {\dfrac{5}{{20}}} \right)}}\Omega$
On dividing the term we get
$= 2 \times \dfrac{1}{{\left( {\dfrac{1}{4}} \right)}}\Omega$
Taking reciprocal of the numerator term and multiply it we get,
$= 8\Omega$

Now,${S_1}$ is closed and ${S_2}$is open
-The diode at ${S_1}$passes current only from C to D or else zero current.
- If current passes through the diode the voltage will be equal to D and C.
-Let us assume that the D and C voltage is equal. Now in the equivalent circuit as $20\Omega {\text{ > }}5\Omega$current has to flow from D to C. From D to C current does not allow. So our assumption is wrong.
We have to treat ${S_1}$as open.
The equivalent resistance is $= \dfrac{{20 + 5}}{2}\Omega$
On adding the numerator term we get
$= \dfrac{{25}}{2}\Omega$
Let us divide the term we get
$= 12.5\Omega$