Certainly! Let's solve **Problem 1 in **2 steps** you requested, with explanations at each and a final answer at the end--- ## ** 1: Maximum and Minimum Current Through the LED** ### **a. Maximum and Minimum Current through the LED** - **Circuit Setup:** The op-amp is configured as a voltage follower, so the voltage across the 10Ω resistor (and LED) is determined by the potentiometer position. - **LED Current Equation:** \[ I_{LED} = \frac{V_{supply} - V_{f}}{R_{total}} \] Where: - \( V_{supply} = 10V \) - \( V_{f} = 2.4V \) (LED forward voltage) - \( R_{total} \) is the sum of the potentiometer setting and the 10Ω resistor. - **Maximum Current:** Potentiometer at **Ω** (minimum resistance). \[ I_{max} = \frac{10V - 2.4V}{10\Omega} = \frac{7.6V}{10\Omega} = .76A \] **Explanation:** Here, all the supply voltage minus the LED drop appears across the 10Ω resistor. - **Minimum Current:** Potentiometer at **100kΩ** (maximum resistance). \[ I_{min} = \frac{10V - 2.4V}{100,010\Omega} \approx \frac{7.6V}{100,010\Omega} \approx .000076A = 76\mu A \] **Explanation:** Most of the voltage is dropped across the potentiometer, resulting in a tiny current. --- ## **Step 2: Maximum Voltage Across the LED, Required Positive Supply, and Power Dissipation** ### **b. Maximum Voltage Across the LED** - The LED always sees its forward voltage, \( V_f = 2.4V \). - **Explanation:** The LED voltage doesn't change regardless of current (idealized LED model). ### **c. Minimum Positive Supply Voltage for Max Current** - To achieve \( I_{LED} = 20mA \) at \( V_f = 2.4V \) (set by the potentiometer): \[ V_{supply} = V_f + (I_{LED} \times 10\Omega) \] \[ V_{supply} = 2.4V + (.02A \times 10\Omega) = 2.4V + .2V = 2.6V \] **Round up to the nearest volt:** **3V** - **Explanation:** This is the minimum supply needed for the op-amp to push 20mA through the LED and resistor. ### **d. Maximum Power Dissipated by the LED and 10Ω Resistor** - **At Maximum Current (.76A):** - **LED Power:** \( P_{LED} = V_f \times I_{max} = 2.4V \times .76A = 1.824W \) - **10Ω Resistor Power:** \( P_{R} = I_{max}^2 \times 10\Omega = (.76A)^2 \times 10\Omega = .5776 \times 10 = 5.776W \) **Explanation:** Power is highest when current is highest, which occurs with the potentiometer at Ω. --- ## **Final Answers Summary** 1. **a.** - **Maximum current:** **.76A** (potentiometer at Ω) - **Minimum current:** **76μA** (potentiometer at 100kΩ) 2. **b.** - **Maximum voltage across LED:** **2.4V** 3. **c.** - **Minimum positive supply for 20mA (rounded):** **3V** 4. **d.** - **Maximum power dissipated by LED:** **1.824W** - **Maximum power dissipated by resistor:** **5.776W** Let me know if you want the next problem solved in the same way!