1. Problem: Determine $V_L$, $I_L$, $I_Z$, and $I_R$ for the given Zener diode circuit with $R_L=180\ \Omega$. Given: $V_Z=10$ V, $P_{Zmax}=400$ mW, $R=220\ \Omega$, $V_{in}=20$ V. 2. Use Ohm's law and power formulas. The Zener voltage $V_Z$ is constant at 10 V when the diode is on. 3. Calculate load current $I_L=\frac{V_L}{R_L}=\frac{10}{180}=0.0556$ A. 4. Calculate total current through resistor $R$: $I_R=\frac{V_{in}-V_Z}{R}=\frac{20-10}{220}=0.0455$ A. 5. Calculate Zener current $I_Z=I_R - I_L=0.0455 - 0.0556 = -0.0101$ A (negative means Zener is off, so $I_Z=0$ and $V_L$ will be less than $V_Z$). 6. Since $I_Z$ cannot be negative, recalculate $V_L$ assuming Zener off: $V_L=V_{in} - I_R R$ with $I_R=\frac{V_L}{R_L}$. 7. Solve $V_L=V_{in} - \frac{V_L}{R_L} R$ \Rightarrow $V_L (1 + \frac{R}{R_L})=V_{in}$ \Rightarrow $V_L=\frac{V_{in}}{1 + \frac{R}{R_L}}=\frac{20}{1 + \frac{220}{180}}=\frac{20}{1 + 1.222}=\frac{20}{2.222}=9$ V. 8. Now $I_L=\frac{9}{180}=0.05$ A, $I_R=\frac{20-9}{220}=\frac{11}{220}=0.05$ A, $I_Z=I_R - I_L=0$ A. 9. For $R_L=470\ \Omega$, repeat steps: $I_L=\frac{V_L}{470}$, $I_R=\frac{20-10}{220}=0.0455$ A. Assuming Zener on, $I_L=0.0455$ A, $V_L=I_L \times 470=0.0455 \times 470=21.4$ V which is impossible (greater than $V_Z$). Assuming Zener off, solve $V_L=\frac{20}{1 + \frac{220}{470}}=\frac{20}{1 + 0.468}=\frac{20}{1.468}=13.63$ V. $I_L=\frac{13.63}{470}=0.029$ A, $I_R=\frac{20-13.63}{220}=0.028$ A, $I_Z=I_R - I_L= -0.001$ A (Zener off). 10. Maximum power condition for Zener diode occurs when $I_Z$ is maximum without exceeding $P_{Zmax}$. Calculate max $I_Z=\frac{P_{Zmax}}{V_Z}=\frac{0.4}{10}=0.04$ A. 11. Total current $I_R=I_L + I_Z=\frac{V_Z}{R_L} + 0.04$. Using $I_R=\frac{V_{in} - V_Z}{R}$, solve for $R_L$: $\frac{V_{in} - V_Z}{R} = \frac{V_Z}{R_L} + 0.04$. Rearranged: $\frac{V_Z}{R_L} = \frac{V_{in} - V_Z}{R} - 0.04$. $R_L = \frac{V_Z}{\frac{V_{in} - V_Z}{R} - 0.04} = \frac{10}{\frac{10}{220} - 0.04} = \frac{10}{0.0455 - 0.04} = \frac{10}{0.0055} = 1818.18\ \Omega$. 12. Minimum $R_L$ to keep Zener on is when $I_Z=0$: $\frac{V_{in} - V_Z}{R} = \frac{V_Z}{R_L}$. $R_L = \frac{V_Z R}{V_{in} - V_Z} = \frac{10 \times 220}{10} = 220\ \Omega$. Final answers: - For $R_L=180\ \Omega$: $V_L=9$ V, $I_L=0.05$ A, $I_Z=0$ A, $I_R=0.05$ A. - For $R_L=470\ \Omega$: $V_L=13.63$ V, $I_L=0.029$ A, $I_Z=0$ A, $I_R=0.028$ A. - Maximum power condition $R_L=1818.18\ \Omega$. - Minimum $R_L$ to keep Zener on $R_L=220\ \Omega$.