Question:
In a switching circuit, if the combination \((S_1∧S_2)\) is connected in parallel to the combination \((S_1^'∧S_2^')\), then the room is lit only when \(\ldots\)
In a switching circuit, if the combination \((S_1∧S_2)\) is connected in parallel to the combination \((S_1^'∧S_2^')\), then the room is lit only when \(\ldots\)
Show Hint
(S1 and S2) in parallel with (S1' and S2') means S1 and S2 are in the same state.
Updated On: Oct 1, 2026
- \(S_1\) is ON and \(S_2\) is OFF
- \(S_1\) is OFF and \(S_2\) is ON
- \(S_1\) and \(S_2\) both ON or \(S_1\) and \(S_2\) both OFF
- The room is always lit.
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Switches in series correspond to \(\wedge\) (AND) and switches in parallel correspond to \(\vee\) (OR). The room is lit when the circuit has at least one closed path.
Step 2: Key Formula or Approach:
The circuit expression is \((S_1 \wedge S_2) \vee (S_1' \wedge S_2')\).
\(S_1'\) is ON exactly when \(S_1\) is OFF.
Step 3: Detailed Explanation:
The first branch conducts when both \(S_1\) and \(S_2\) are ON.
The second branch conducts when \(S_1'\) and \(S_2'\) are both ON, which means both \(S_1\) and \(S_2\) are OFF.
So the lamp is lit when both are ON or both are OFF. This is the biconditional \(S_1 \leftrightarrow S_2\).
If \(S_1\) is ON and \(S_2\) is OFF (option A), the first branch is broken at \(S_2\) and the second branch is broken at \(S_1'\), so no path exists. The same holds for option (B). The room is not always lit, so (D) is wrong.
Final Answer:
The room is lit when both switches are ON or both are OFF, option (C). \[ \boxed{\text{Both ON or both OFF (C)}} \]
Switches in series correspond to \(\wedge\) (AND) and switches in parallel correspond to \(\vee\) (OR). The room is lit when the circuit has at least one closed path.
Step 2: Key Formula or Approach:
The circuit expression is \((S_1 \wedge S_2) \vee (S_1' \wedge S_2')\).
\(S_1'\) is ON exactly when \(S_1\) is OFF.
Step 3: Detailed Explanation:
The first branch conducts when both \(S_1\) and \(S_2\) are ON.
The second branch conducts when \(S_1'\) and \(S_2'\) are both ON, which means both \(S_1\) and \(S_2\) are OFF.
So the lamp is lit when both are ON or both are OFF. This is the biconditional \(S_1 \leftrightarrow S_2\).
If \(S_1\) is ON and \(S_2\) is OFF (option A), the first branch is broken at \(S_2\) and the second branch is broken at \(S_1'\), so no path exists. The same holds for option (B). The room is not always lit, so (D) is wrong.
Final Answer:
The room is lit when both switches are ON or both are OFF, option (C). \[ \boxed{\text{Both ON or both OFF (C)}} \]
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