Let's break down each part step by step and clarify what is being asked and how you should approach it. --- ## (a) The Norton Equivalent Current **Task:** Find the Norton equivalent current (\( I_{Norton} \)) for the circuit at terminals AB, assuming \( I_x = \). ### Steps: 1. **Remove the load:** Consider the circuit open at AB. 2. **Set \( I_x = \):** The current source is inactive (an open circuit). 3. **Find the current through a short across AB:** To get the Norton current, short AB and calculate the current through the short. 4. **Simplify the circuit:** - Combine \( 7k \Omega \) and \( 11k \Omega \) in parallel (since both connect from A to B). - The left-hand side forms a voltage divider with the \( 5V \) source. 5. **Find the equivalent resistance and the current:** - Calculate the voltage at A with respect to B (find open-circuit voltage). - Find the resistance seen by the short between A and B. - Use Ohm's law to find the Norton current. **Let me know if you want a step-by-step calculation here.** --- ## (b) Value of \( I_x \) to Halve the Thevenin Voltage **Task:** Determine the value of \( I_x \) that causes the Thevenin voltage (open-circuit voltage at AB) to become half of what it was when \( I_x = \). ### Steps: 1. **Calculate Thevenin voltage for \( I_x = \):** - This is the open-circuit voltage at AB from part (a). 2. **Express Thevenin voltage as a function of \( I_x \):** - With \( I_x \) active, it injects current from A to B. - The open-circuit voltage at AB will be the sum of the voltage from the left-side network and the voltage drop across the parallel combination of \( 7k \) and \( 11k \) due to \( I_x \). 3. **Set up the equation:** - \( V_{th, new} = \frac{1}{2} V_{th, old} \) - Solve for \( I_x \). --- ## (c) Value of \( R_{load} \) for 25% Power Consumption **Task:** Find \( R_{load} \) such that it consumes 25% of the total power delivered by the circuit for the \( I_x \) found in part (b). ### Steps: 1. **Total Power Delivered:** - The total power is the power delivered by the Thevenin equivalent to \( R_{load} \). 2. **Fraction Consumed by Load:** - \( P_{load} = .25 \times P_{total} \) 3. **Set up the equation:** - Power delivered to \( R_{load} \): \( P_{load} = \frac{V_{th}^2}{(R_{th} + R_{load})^2} R_{load} \) - Total power: \( P_{total} = \frac{V_{th}^2}{R_{th}} \) (if only \( R_{th} \)), but since both load and internal resistance are in series, the total power from the source is split across both. - The fraction is: \[ \frac{P_{load}}{P_{total}} = \frac{R_{load}}{R_{th} + R_{load}} \] Set equal to .25 and solve for \( R_{load} \). --- ## **Summary Table** | Part | What to Find | Key Steps | |------|--------------|-----------| | a | Norton equivalent current | Short AB, find current with \( I_x = \) | | b | Value of \( I_x \) to halve Thevenin voltage | Set up \( V_{th,new} = .5 V_{th,old} \), solve for \( I_x \) | | c | Value of \( R_{load} \) for 25% power | Solve \( \frac{R_{load}}{R_{th} + R_{load}} = .25 \) | --- ## **Do you need full calculations for each part, or is this conceptual breakdown enough to help you get started?**

Sure! Let's work through each part of the question with detailed calculations. --- ## (a) The Norton Equivalent Current ### Given Circuit Parameters: - Resistors: \( R_1 = 3k\Omega \), \( R_2 = 11k\Omega \), \( R_3 = 7k\Omega \) - Voltage Source: \( V = 5V \) ### Steps: 1. **Remove the load and short terminals AB:** We will calculate the Norton equivalent current by shorting the terminals AB and finding the current flowing through the short. 2. **Calculate the equivalent resistance:** The resistors \( R_2 \) and \( R_3 \) are in parallel: \[ R_{eq} = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{11k \cdot 7k}{11k + 7k} = \frac{77}{18} k\Omega \approx 4.28k\Omega \] 3. **Calculate the total current from the voltage source:** Using Ohm's law, the current from the voltage source is: \[ I = \frac{V}{R_{eq} + R_1} = \frac{5V}{4.28k + 3k} \approx \frac{5}{7.28} mA \approx 0.686 mA \] 4. **Calculate the Norton current:** The current flowing through the short (Norton current) is the same as the total current from the source: \[ I_{Norton} = 0.686 mA \] ### Result: \[ \text{The Norton equivalent current } I_{Norton} = 0.686 mA \] --- ## (b) Value of \( I_x \) to Halve the Thevenin Voltage ### Steps: 1. **Calculate Thevenin voltage \( V_{th} \) when \( I_x = 0 \):** Using voltage division: \[ V_{th} = \frac{R_3}{R_2 + R_3} \cdot V = \frac{7k}{11k + 7k} \cdot 5V = \frac{7}{18} \cdot 5V \approx 1.94V \] 2. **Set \( V_{th, new} = \frac{1}{2} V_{th, old} \):** \[ V_{th, new} = \frac{1}{2} \cdot 1.94V = 0.97V \] 3. **Express \( V_{th} \) as a function of \( I_x \):** The voltage across \( R_3 \) when \( I_x \) is active: \[ V_{AB} = V - I_x \cdot R_3 \] 4. **Set up the equation:** \[ \frac{7}{18} \cdot 5V - I_x \cdot 7k = 0.97V \] Rearranging gives: \[ I_x \cdot 7k = 1.94V - 0.97V \implies I_x \cdot 7k = 0.97V \implies I_x = \frac{0.97V}{7k} \approx 0.139 mA \] ### Result: \[ \text{The value of } I_x \text{ to halve the Thevenin voltage is } 0.139 mA. \] --- ## (c) Value of \( R_{load} \) for 25% Power Consumption ### Steps: 1. **Total power delivered by the Thevenin equivalent:** \[ P_{total} = \frac{V_{th}^2}{R_{th}} = \frac{(1.94V)^2}{4.28k\Omega} \approx \frac{3.76}{4.28} mW \approx 0.876 mW \] 2. **Calculate power consumed by the load:** To consume 25% of the total power: \[ P_{load} = 0.25 \cdot P_{total} = 0.25 \cdot 0.876 mW \approx 0.219 mW \] 3. **Power delivered to \( R_{load} \):** \[ P_{load} = \frac{V_{th}^2}{(R_{th} + R_{load})^2} R_{load} \] 4. **Setting up the equation for power consumption:** \[ 0.219 mW = \frac{(1.94V)^2}{(4.28k + R_{load})^2} R_{load} \] Rearranging and solving becomes complex, so we can assume \( R_{load} \) is in the same range and use iteration or numerical methods to find \( R_{load} \). ### Approximation: Assuming \( R_{load} \approx 4.28k\Omega \): \[ R_{load} \approx 1.42k\Omega \] ### Result: \[ R_{load} \approx 1.42k\Omega \text{ ensures } R_{eq} \text{ only consumes 25% of the total power delivered.} \] --- ## **Final Summary:** 1. **Norton Current:** \( I_{Norton} = 0.686 mA \) 2. **Value of \( I_x \):** \( I_x = 0.139 mA \) 3. **Value of \( R_{load} \):** \( R_{load} \approx 1.42k\Omega \) This completes the solution with detailed calculations! Let me know if you need any further assistance.