### Related Formula I_texttotal = I_Z + I_L$I_{\text{total}} = I_Z + I_L$ I_texttotal = fracV_textin - V_ZR_S$I_{\text{total}} = \frac{V_{\text{in}} - V_Z}{R_S}$ R_L = fracV_ZI_L$R_L = \frac{V_Z}{I_L}$ ### Core Logic Given parameters: - Input voltage, V_textin = 25mathrm~V$V_{\text{in}} = 25\mathrm{~V}$ - Zener breakdown voltage, V_Z = 5mathrm~V$V_Z = 5\mathrm{~V}$ - Series resistor, R_S = 400mathrm~Omega$R_S = 400\mathrm{~\Omega}$ The voltage drop across the series resistor R_S$R_S$ is: V_R_S = V_textin - V_Z = 25 - 5 = 20mathrm~V$V_{R_S} = V_{\text{in}} - V_Z = 25 - 5 = 20\mathrm{~V}$ The total series current I$I$ is: I_texttotal = fracV_R_SR_S = frac20400 = 0.05mathrm~A = 50mathrm~mA$I_{\text{total}} = \frac{V_{R_S}}{R_S} = \frac{20}{400} = 0.05\mathrm{~A} = 50\mathrm{~mA}$ We are given that the zener current I_Z$I_Z$ is 4 times the load current I_L$I_L$: I_Z = 4 I_L$I_Z = 4 I_L$ Since I_texttotal = I_Z + I_L$I_{\text{total}} = I_Z + I_L$: 50mathrm~mA = 4 I_L + I_L = 5 I_L implies I_L = 10mathrm~mA$50\mathrm{~mA} = 4 I_L + I_L = 5 I_L \implies I_L = 10\mathrm{~mA}$ The load resistance connected in parallel with the Zener diode is: R_L = fracV_ZI_L = frac5mathrm~V10 times 10^-3mathrm~A = 500mathrm~Omega$R_L = \frac{V_Z}{I_L} = \frac{5\mathrm{~V}}{10 \times 10^{-3}\mathrm{~A}} = 500\mathrm{~\Omega}$ ### Step 1: Final Conclusion The load current is 10mathrm~mA$10\mathrm{~mA}$ and the load resistance is 500mathrm~Omega$500\mathrm{~\Omega}$. ### Pattern Recognition In any zener regulator problem: first determine the series path current using I_texttotal = fracV_textin - V_ZR_s$I_{\text{total}} = \frac{V_{\text{in}} - V_Z}{R_s}$. Split this total current using the given ratio of I_Z$I_Z$ and I_L$I_L$. Finally, determine R_L$R_L$ from V_Z / I_L$V_Z / I_L$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### 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