Solve the Following Question.(4 Marks)

Question 14 Marks

Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.

Answer

(ii) Let $p :$ the switch $S_1$ is closed
$q :$ the switch $S_2$ is closed
$r :$ the switch $S_3$ is closed
$s :$ the switch $S_4$ is closed
$t :$ the switch $S_5$ is closed
$\sim p :$ the switch $S_1‘$ is closed or the switch $S_1$ is open
$\sim q :$ the switch $S_2‘$ is closed or the switch $S_2$ is open
$\sim r :$ the switch $S_3‘$ is closed or the switch $S_3$ is open
$\sim s :$ the switch $S_4‘$ is closed or the switch $S_4$ is open
$\sim t :$ the switch $S_5‘$ is closed or the switch $S_5$ is open.
Then the given circuit in symbolic form is
$[(p ∧ q) ∨ \sim r ∨ \sim s ∨ \sim t] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)]$
Using the laws of logic, we have,
$[(p ∧ q) ∨ \sim r ∨ \sim s ∨ \sim t] ∧ [(p A q) ∨ (r ∧ s ∧ t)]$
$= [(p∧ q) ∨ \sim (r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] … ($By De Morgan’s Law$)$
$= (p ∧ q) ∨ [ \sim (r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] … ($By Distributive Law$)$
$= (p ∧ q) ∨ F … ($By Complement Law$)$
$= p ∧ q … ($By Identity Law$)$
Hence, the alternative simplified circuit is :

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Question 24 Marks

Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.

Answer

Let $p :$ the switch $S_1$ is closed
$q :$ the switch $S_2$ is closed
$\sim p:$ the switch $S_1‘$ is closed or the switch $S_1$ is open
$\sim q:$ the switch $S_2‘$ is closed or the switch $S_2$ is open.
Then the given circuit in symbolic form is :
$(p ∧ \sim q) ∨ (\sim p ∧ q) ∨ (\sim p ∧ \sim q)$
Using the laws of logic, we have,
$(p ∧ \sim q) ∨ (\sim p ∧ q) ∨ (\sim p ∧ \sim q)$
$= (p ∧ \sim q) ∨ [(\sim p ∧ q) ∨ (\sim p ∧ \sim q) …($By Complement Law$)$
$= (p ∧ \sim q) ∨ [\sim p ∧ (q ∨ \sim q) ($By Distributive Law$)$
$= (p ∧ \sim q) ∨ (\sim p ∧ T) …($By Complement Law$)$
$= (p ∧ \sim q) ∨ \sim p …($By Identity Law$)$
$= (p ∨ \sim p) ∧ (\sim q ∨ \sim p) …($By Distributive Law$)$
$= \sim q ∨ \sim p …($By Identity Law$)$
$= \sim p ∨ \sim p …($By Commutative Law$)$
Hence, the simplified circuit for the given circuit is :

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Question 34 Marks

Write the symbolic form of the following switching circuits construct its switching table and interpret it.

Answer

Let $p$ : the switch $S_1$ is closed
$q$ : the switch $S_2$ is closed
$r$ : the switch $S_3$ is closed
$~q$ : the switch $S_2$‘ is closed or the switch $S_2$ is open
$~r$: the switch $S_3$‘ is closed or the switch $S_3$ is open.
Then the symbolic form of the given circuit is :$[p ∨ (~q) ∨ r)] ∧ [p ∨ (q ∧ r)]$

From the switching table, the ‘final column’ and the column of p are identical. Hence, the lamp will glow which $S_1$ is $‘ON’$.

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Question 44 Marks

Write the symbolic form of the following switching circuits construct its switching table and interpret it.

Answer

Let $p :$ the switch $S_1$ is closed
$q :$ the switch $S_2$ is closed
$\sim p :$ the switch $S_1$ is closed or the switch $S_1$ is open.
$\sim q :$ the switch $S_2‘$ is closed or the switch $S_2$ is open.
Then the symbolic form of the given circuit is $: p ∨ (\sim p ∧ \sim q) ∨ (p ∧ q)$

Since the final column contains $‘0’$ when $p$ is $0$ and $q$ is $‘1’$, otherwise it contains $‘1′.$
Hence, the lamp will not glow when $S_1$ is OFF and $S_2$ is ON, otherwise the lamp will glow.

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Question 54 Marks

Write the symbolic form of the following switching circuits construct its switching table and interpret it.

Answer

Let $p :$ the switch $S_1$ is closed
$q :$ the switch $S_2$ is closed
$\sim p :$ the switch $S_1‘$ is closed or the switch $S_1$ is open
$\sim q :$ the switch $S_2‘$ is closed or the switch $S_2$ is open.
Then the symbolic form of the given circuit is :
$(p ∨ \sim q) ∨ (\sim p ∧ q)$

Since the final column contains all’ $1′,$ the lamp will always glow irrespective of the status of switches.

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Maths Solve the Following Question.(4 Marks)

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