Logic Circuit Inputs for Glowing Both LEDs

To determine the combination of inputs $(A, B, C)$ that will cause both LED-1 and LED-2 to glow, we analyze the circuit stage-by-stage:

1. Identify the Logic Gates in the Circuit

• Gate 1 (Leftmost): An OR gate with inputs $A$ and $B$.

  • Output of Gate 1: $Y_1 = A + B$
  • LED-1 is connected to the output of Gate 1. Therefore, for LED-1 to glow, we require: $Y_1 = 1 \implies A + B = 1$

• Gate 2 (Middle): An AND gate with inputs $Y_1$ and $C$.

  • Output of Gate 2: $Y_2 = Y_1 \cdot C = (A + B) \cdot C$

• Gate 3 (Rightmost): An AND gate with inputs $A$ and $Y_2$.

  • Output of Gate 3: $Y_3 = A \cdot Y_2 = A \cdot (A + B) \cdot C$
  • LED-2 is connected to the output of Gate 3. Therefore, for LED-2 to glow, we require: $Y_3 = 1 \implies A \cdot (A + B) \cdot C = 1$

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2. Condition for Both LEDs to Glow

For both LED-1 and LED-2 to glow simultaneously, we must have $Y_1 = 1$ and $Y_3 = 1$:

1. Since $Y_3 = A \cdot Y_2 = 1$, all inputs to Gate 3 must be HIGH ($1$): $A = 1 \quad \text{and} \quad Y_2 = 1$

2. Since $Y_2 = Y_1 \cdot C = 1$, all inputs to Gate 2 must be HIGH ($1$): $Y_1 = 1 \quad \text{and} \quad C = 1$

3. Combining these conditions, we must have: $A = 1 \quad \text{and} \quad C = 1$

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3. Verification of the Given Options

• Option A ($A = 0, B = 1, C = 1$):

  • $Y_3 = 0 \cdot (0 + 1) \cdot 1 = 0 \implies$ LED-2 does not glow.

• Option B ($A = 1, B = 0, C = 0$):

  • $Y_3 = 1 \cdot (1 + 0) \cdot 0 = 0 \implies$ LED-2 does not glow.

• Option C ($A = 1, B = 0, C = 1$):

  • $Y_1 = 1 + 0 = 1 \implies$ LED-1 glows.
  • $Y_2 = 1 \cdot 1 = 1$
  • $Y_3 = 1 \cdot 1 = 1 \implies$ LED-2 glows.

• Option D ($A = 1, B = 1, C = 0$):

  • $Y_3 = 1 \cdot (1 + 1) \cdot 0 = 0 \implies$ LED-2 does not glow.

Thus, the correct combination of inputs is $A = 1, B = 0, C = 1$.