Q.
The switching function for switching network is
A) $\left( x \land y^{'} \right) \vee \left( y \land \mathcal{z}^{'} \right)\mathcal{\vee (z \land}x^{'})$
B) $\left( x \land y \land \mathcal{z} \right) \vee (x^{'} \land y^{'} \land \mathcal{z}^{'})$
C) $\left( x \vee y^{'} \right) \land \left( y \vee \mathcal{z}^{'} \right)\mathcal{\land (z \vee}x^{'})$
D) None of the above
Explanation
(a)
$\because$ Switches $x$ and $y^{'}$ are connected parallel which is denoted by $(x{\land y}^{'})$
Similarly, $y$ and $\mathcal{z}^{'}$and $\mathcal{z}$ and $x^{'}$ are also connected parallel
Which are denoted by $(y \land \mathcal{z}^{'})$ and $\mathcal{(z \land}x^{'})$ respectively
Now, $x \land y^{'},\ y \land \mathcal{z}^{'}$ and $\mathcal{z \land}x^{'}$ are connected in series. So, switching function of given network is
$\left( x \land y^{'} \right) \vee \left( y \land \mathcal{z}^{'} \right)\mathcal{\vee (z \land}x^{'})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.