A zener diode with 5mathrm~V$5\mathrm{~V}$ zener voltage is used to regulate an unregulated dc voltage input of 25mathrm~V$25\mathrm{~V}$. For a 400mathrm~Omega$400\mathrm{~\Omega}$ resistor connected in series, the zener current is found to be 4 times load current. The load current (I_L)$(I_L)$ and load resistance (R_L)$(R_L)$ are:

### Related Formula Boolean operators for basic logic gates: - OR gate: Y = A + B$Y = A + B$ - AND gate: Y = A cdot B$Y = A \cdot B$ ### Core Logic Analyze the schematic: 1. Input lines A$A$ and B$B$ are linked into an OR gate, yielding output: Y = A + B$Y = A + B$ 2. This term A + B$A + B$ and the original input line A$A$ form the inputs to a final AND gate. 3. The final output expression C$C$ is therefore: C = A cdot (A + B)$$C = A \cdot (A + B)$$

### Step 1: Construct the Truth Table Compute output values for each input combination: - For A=0, B=0$A=0, B=0$: C = 0 cdot (0 + 0) = 0$$C = 0 \cdot (0 + 0) = 0$$ - For A=1, B=0$A=1, B=0$: C = 1 cdot (1 + 0) = 1 cdot 1 = 1$$C = 1 \cdot (1 + 0) = 1 \cdot 1 = 1$$ - For A=0, B=1$A=0, B=1$: C = 0 cdot (0 + 1) = 0 cdot 1 = 0$$C = 0 \cdot (0 + 1) = 0 \cdot 1 = 0$$ - For A=1, B=1$A=1, B=1$: C = 1 cdot (1 + 1) = 1 cdot 1 = 1$$C = 1 \cdot (1 + 1) = 1 \cdot 1 = 1$$ This matches Table 2. ### Pattern Recognition Using Boolean algebra: C = A cdot (A + B) = A cdot A + A cdot B = A + A cdot B$$C = A \cdot (A + B) = A \cdot A + A \cdot B = A + A \cdot B$$ By absorption law: A + A cdot B = A$$A + A \cdot B = A$$ So the circuit is equivalent to a simple buffer carrying input A$A$. The output C$C$ must match input A$A$ under all conditions. Checking the options, Table 2 is the one where output C$C$ tracks input A$A$ directly (0, 1, 0, 1$0, 1, 0, 1$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductors

### Core Logic Let the intermediate outputs of the first layers be Y_1$Y_1$ and Y_2$Y_2$ [cite: 747]: * Top gate is an AND gate with inputs A$A$ and B$B$, so Y_1 = A cdot B$Y_1 = A \cdot B$ [cite: 747]. * Bottom gate is an OR gate where one input is B$B$ and the other is barA$\bar{A}$ via a NOT gate, so Y_2 = barA + B$Y_2 = \bar{A} + B$ [cite: 747]. * The final layer is a NAND gate with inputs Y_1$Y_1$ and Y_2$Y_2$, so Y = overlineY_1 cdot Y_2$Y = \overline{Y_1 \cdot Y_2}$[cite: 748]. ### Step 1: Constructing the Truth Table Let's compute the output Y$Y$ for all binary input pairs (A, B)$(A, B)$ [cite: 757]: * For A=0, B=0 implies Y_1 = 0, Y_2 = 1 implies Y = overline0 cdot 1 = 1$A=0, B=0 \implies Y_1 = 0, Y_2 = 1 \implies Y = \overline{0 \cdot 1} = 1$ [cite: 757]. * For A=1, B=0 implies Y_1 = 0, Y_2 = 0 implies Y = overline0 cdot 0 = 1$A=1, B=0 \implies Y_1 = 0, Y_2 = 0 \implies Y = \overline{0 \cdot 0} = 1$ [cite: 757]. * For A=0, B=1 implies Y_1 = 0, Y_2 = 1 implies Y = overline0 cdot 1 = 1$A=0, B=1 \implies Y_1 = 0, Y_2 = 1 \implies Y = \overline{0 \cdot 1} = 1$ [cite: 757]. * For A=1, B=1 implies Y_1 = 1, Y_2 = 1 implies Y = overline1 cdot 1 = 0$A=1, B=1 \implies Y_1 = 1, Y_2 = 1 \implies Y = \overline{1 \cdot 1} = 0$ [cite: 757]. Thus, Y=0$Y=0$ uniquely when A=1$A=1$ and B=1$B=1$[cite: 89, 90, 757]. ### Pattern Recognition A NAND gate produces an output of 0$0$ if and only if all its inputs are 1$1$. Working backward, this instantly sets Y_1=1$Y_1=1$ and Y_2=1$Y_2=1$. For Y_1 = A cdot B = 1$Y_1 = A \cdot B = 1$, we must have A=1$A=1$ and B=1$B=1$ simultaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductors

### Related Formula The Boolean algebraic relationships for standard gates are: * AND Gate: Y = A cdot B$Y = A \cdot B$ * NOT Gate: Y = barA$Y = \bar{A}$ * OR Gate: Y = A + B$Y = A + B$ ### Core Logic Analyzing the connections of the logic circuit shown in the diagram: 1. Input A$A$ is passed directly to the top AND gate. 2. Input A$A$ is also passed through a NOT gate, giving barA$\bar{A}$, which is supplied to the bottom AND gate. 3. Input B$B$ is connected directly to both the top and bottom AND gates. Hence: * Output of the top AND gate is: A cdot B$A \cdot B$ * Output of the bottom AND gate is: barA cdot B$\bar{A} \cdot B$ Both outputs are then combined by an OR gate to give the final output Y$Y$: Y = (A cdot B) + (barA cdot B)$$Y = (A \cdot B) + (\bar{A} \cdot B)$$ ### Step 1: Simplify the Boolean Expression Using Boolean algebra, factor out B$B$: Y = (A + barA) cdot B$$Y = (A + \bar{A}) \cdot B$$ Since A + barA = 1$A + \bar{A} = 1$: Y = 1 cdot B$Y = 1 \cdot B$ Y = B$Y = B$

### Step 2: Construct the Truth Table Because the output Y$Y$ is functionally identical to B$B$, the truth table columns for B$B$ and Y$Y$ must be exactly the same: * When B = 0$B = 0$, Y = 0$Y = 0$ * When B = 1$B = 1$, Y = 1$Y = 1$ This yields the matching truth table in Option 2. ### Pattern Recognition Observe: Y = A cdot B + barA cdot B$Y = A \cdot B + \bar{A} \cdot B$. Whenever B$B$ is a common factor to both paths of the AND-OR configuration, it suggests the expression can be simplified directly to B$B$ because the state of A$A$ becomes irrelevant (A + barA = 1$A + \bar{A} = 1$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductors