 Hello and welcome to the session. In this session we discuss the following question which says find the Boolean function of the switching circuit given here and then its equivalent simplified circuit also draw the simplified circuit. Let's proceed with the solution now. This is the switching circuit given to us and we will first find the Boolean function for this switching circuit. In this switching circuit we have this B as the battery, L as the light, P and Q complements are the switches. In this upper branch then in the middle branch we have three switches P, Q and R and in the lower branch there are two switches P complement and R. So, from the figure we observe that in the upper branch P and Q complement are two switches in series and so it is represented by the function P into Q complement. Now we move on to the middle branch. In this middle branch there are three switches P, Q and R and they are also in series and so it would be represented by the function P into Q into R. Then let's see the lower branch. In the lower branch there are two switches P complement and R. They are in series so they would be represented by the function P complement into R. Now from the switching circuit we observe all three branches are in parallel. Therefore the whole circuit is represented by the function P into Q complement which is for the upper branch plus P, Q, R which is for the middle branch plus P complement into R which is for the lower branch. And so the required Boolean function is P into Q complement plus P into Q into R plus P complement into R. Now we will reduce this Boolean function to a simplified circuit by using the laws of Boolean algebra. So we consider the Boolean function P into Q complement plus P into Q into R plus P complement into R. For this Boolean algebra we have a distributive law according to which we have A into B plus C the whole is equal to A into B plus A into C. This is the distributive law. Now we will use this distributive law for this expression. So now here taking P as A, Q complement as B and Q R as C and applying the distributive law here we get this is equal to P into Q complement plus Q R the whole plus P complement into R. We have another distributive law A plus B into C the whole is equal to A plus B the whole into A plus C the whole. So we apply this distributive law here that is for the expression Q complement plus Q R. So this is further equal to P into Q complement plus Q the whole into Q complement plus R the whole. This whole plus P complement R. Now for any element A there exists its inverse A complement such that A plus A complement is equal to one which is same as A complement plus A where this one is the identity element for the operation of product. So Q complement plus Q could be written as one so this is equal to P into one into Q complement plus R the whole this whole plus P complement R. So here we have used Q complement plus Q is equal to one. Further we have A into one is equal to A which is same as one into A where this one is the identity element for the operation of the product. So this is further equal to P into now in place of one into Q complement plus R the whole. We can write Q complement plus R the whole this whole plus P complement R. Now we can apply this distributive law for this expression. So applying the distributive law we have this is equal to P into Q complement plus P into R plus P complement into R. This expression P into R plus P complement into R we can again apply the distributive law. So we get this is further equal to P into Q complement plus P complement the whole into R. Now P plus P complement would be equal to one so this is equal to P into Q complement plus one into R. Now again one into R is same as R so P into Q complement plus R. So the given Boolean function was P into Q complement plus P Q R plus P complement into R. We have simplified this Boolean function as P into Q complement plus R. Therefore P into Q complement plus R is the simplified circuit equivalent to the given circuit. Now we will draw this circuit that is P into Q complement plus R. So we have drawn this circuit P into Q complement plus R. Now P into Q complement means that the switches P and Q complement are in series which we have shown here in the upper branch. And in the lower branch we have the switch R which is in parallel to the series of the switches P and Q complement. So this completes the session hope you have understood the solution of this question.