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If the symbolic form of the switching circuit is $[\sim \mathrm{pv}(\mathrm{p} \wedge \sim \mathrm{q})] \mathrm{v} \mathrm{q}$, then the current flows through the circuit onlyif
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The correct answer is:
irrespective of status of the switches
$[\sim p \vee(p \wedge \sim q)] \vee q$
$=[(\sim p \vee p) \wedge(\sim p \vee \sim q)] \vee q$
$E[T \wedge(\sim p \vee \sim q)] \vee q$
$\equiv(\sim p \vee \sim q) \vee q$
$\equiv \sim p \vee(\sim q \vee q) \equiv \sim p \vee T \equiv T$
This shows that current flows irrespective of status of the switches.
$=[(\sim p \vee p) \wedge(\sim p \vee \sim q)] \vee q$
$E[T \wedge(\sim p \vee \sim q)] \vee q$
$\equiv(\sim p \vee \sim q) \vee q$
$\equiv \sim p \vee(\sim q \vee q) \equiv \sim p \vee T \equiv T$
This shows that current flows irrespective of status of the switches.
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