If p: switch s_1 is closed,q: switch s_2 is c | vedclass

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(B) The circuit consists of switch $s_1$ in series with a parallel combination of $s_1'$ and $s_2'$,which is then in series with $s_2$.Symbolically,th

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The lamp is always off

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(B) The circuit consists of switch $s_1$ in series with a parallel combination of $s_1'$ and $s_2'$,which is then in series with $s_2$.Symbolically,this is represented as $p \wedge (\sim p \vee \sim q) \wedge q$.Using the distributive law: $p \wedge ((\sim p \vee \sim q) \wedge q) = p \wedge ((\sim p \wedge q) \vee (\sim q \wedge q))$.Since $(\sim q \wedge q)$ is a contradiction $(F)$,we have $p \wedge ((\sim p \wedge q) \vee F) = p \wedge (\sim p \wedge q)$.By associative law: $(p \wedge \sim p) \wedge q = F \wedge q = F$.Since the symbolic form simplifies to a contradiction $(F)$,the lamp is always off.

If $p$: switch $s_1$ is closed,$q$: switch $s_2$ is closed,then the correct interpretation of the following circuit is:

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