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(C) To make both $LEDs$ glow, the output of the gates connected to them must be high $(1)$.Analyzing the circuit diagram:$1$. $LED-1$ is connected to

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$A=1, B=0, C=1$

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(C) To make both $LEDs$ glow, the output of the gates connected to them must be high $(1)$.Analyzing the circuit diagram:$1$. $LED-1$ is connected to the output of the $OR$ gate. Let the output of the $OR$ gate be $Y_1 = A + B$. For $LED-1$ to glow, $Y_1$ must be $1$.$2$. $LED-2$ is connected to the output of the final $AND$ gate. The inputs to this $AND$ gate are the output of the $OR$ gate $(Y_1)$ and input $A$. Thus, the output $Y_2 = Y_1 \cdot A = (A + B) \cdot A$. However, looking closely at the diagram, the inputs to the final $AND$ gate are the output of the middle $AND$ gate and input $A$. The middle $AND$ gate has inputs $Y_1$ and $C$. So, $Y_{middle} = (A + B) \cdot C$. The final $AND$ gate has inputs $Y_{middle}$ and $A$. Thus, $Y_{LED2} = ((A + B) \cdot C) \cdot A$.$3$. For $LED-1$ to glow, $A + B = 1$.$4$. For $LED-2$ to glow, $(A + B) \cdot C \cdot A = 1$. This requires $A=1$, $C=1$, and $(A+B)=1$. Since $A=1$, the condition $(A+B)=1$ is automatically satisfied regardless of $B$.$5$. Checking the options:- For $A=1, B=0, C=1$: $Y_{LED1} = 1+0 = 1$ (Glows), $Y_{LED2} = (1+0) \cdot 1 \cdot 1 = 1$ (Glows).Therefore, option $C$ is correct.

For the given logic circuit, which of the following input combinations will make both $LED-1$ and $LED-2$ glow?

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