Let's work through each part step by step, based on the circuit and the characteristic curves for \( Z_S, Z_L, \) and \( Z_D \). --- ### **1) Draw a graph of the \( I_S \) value according to the \( V_E \) value.** #### **Step 1: Analyze each element’s characteristic curve:** - **\( Z_S \):** \( I \) increases linearly from 0 to 5 as \( V \) goes from 0 to 2. Slope = \( \frac{5}{2} = 2.5 \, \text{A/V} \). - **\( Z_L \):** \( I \) increases linearly from 0 to 4 as \( V \) goes from 0 to 1, then stays at 4 for \( V > 1 \). - **\( Z_D \):** For \( V > 4 \), \( I = 10(V - 4) \). For \( V \leq 4 \), \( I = 0 \). #### **Step 2: Find \( I_S \) as a function of \( V_E \):** - \( I_S = I_{Z_L} + I_{Z_D} \) (since these are in parallel across \( V_E \)) #### **Piecewise calculation:** - **For \( 0 \leq V_E < 1 \):** - \( I_{Z_L} = 4V_E \) - \( I_{Z_D} = 0 \) - \( I_S = 4V_E \) - **For \( 1 \leq V_E < 4 \):** - \( I_{Z_L} = 4 \) (saturated) - \( I_{Z_D} = 0 \) - \( I_S = 4 \) - **For \( V_E \geq 4 \):** - \( I_{Z_L} = 4 \) - \( I_{Z_D} = 10(V_E - 4) \) - \( I_S = 4 + 10(V_E - 4) \) #### **Graph Sketch:** - \( 0 \leq V_E < 1 \): line from (0,0) to (1,4), slope = 4 - \( 1 \leq V_E < 4 \): horizontal line at \( I_S = 4 \) - \( V_E \geq 4 \): slope = 10, line starts at (4,4), so at \( V_E \), \( I_S = 4 + 10(V_E-4) \) --- ### **2) Power consumed in \( Z_L \) (\( P_L \)) as a function of \( V_E \) and its graph** \( P_L = V_E \cdot I_{Z_L} \) - **For \( 0 \leq V_E < 1 \):** - \( I_{Z_L} = 4V_E \) - \( P_L = V_E \cdot 4V_E = 4V_E^2 \) - **For \( V_E \geq 1 \):** - \( I_{Z_L} = 4 \) - \( P_L = 4V_E \) #### **Graph Sketch:** - \( 0 \leq V_E < 1 \): parabola from (0,0) to (1,4) - \( V_E \geq 1 \): straight line with slope 4, starting from (1,4) --- ### **3) Find where \( \frac{P_L}{P_E} \) is maximum** - \( P_E = V_E \cdot I_S \) - \( \frac{P_L}{P_E} = \frac{P_L}{V_E \cdot I_S} \) #### **Calculate for each region:** **Region 1: \( 0 \leq V_E < 1 \)** - \( I_S = 4V_E \) - \( P_L = 4V_E^2 \) - \( P_E = V_E \cdot 4V_E = 4V_E^2 \) - \( \frac{P_L}{P_E} = 1 \) **Region 2: \( 1 \leq V_E < 4 \)** - \( I_S = 4 \) - \( P_L = 4V_E \) - \( P_E = V_E \cdot 4 = 4V_E \) - \( \frac{P_L}{P_E} = 1 \) **Region 3: \( V_E \geq 4 \)** - \( I_S = 4 + 10(V_E - 4) \) - \( P_L = 4V_E \) - \( P_E = V_E \cdot [4 + 10(V_E - 4)] = 4V_E + 10V_E(V_E - 4) \) - \( \frac{P_L}{P_E} = \frac{4V_E}{4V_E + 10V_E(V_E - 4)} = \frac{4}{4 + 10(V_E - 4)} \) - This ratio **decreases** as \( V_E \) increases beyond 4. **Maximum ratio:** - For \( V_E < 4 \), \( \frac{P_L}{P_E} = 1 \) - For \( V_E \geq 4 \), \( \frac{P_L}{P_E} < 1 \) **So the maximum ratio is 1, and it occurs for \( 0 < V_E < 4 \).** --- ## **Summary Table** | Region | \( V_E \) Range | \( I_S \) | \( P_L \) | \( P_E \) | \( \frac{P_L}{P_E} \) | |--------------------|------------------------|---------------------|-------------------|-----------------|----------------------| | 1 | \( 0 \leq V_E < 1 \) | \( 4V_E \) | \( 4V_E^2 \) | \( 4V_E^2 \) | 1 | | 2 | \( 1 \leq V_E < 4 \) | 4 | \( 4V_E \) | \( 4V_E \) | 1 | | 3 | \( V_E \geq 4 \) | \( 4+10(V_E-4) \) | \( 4V_E \) | \( 4V_E+10V_E(V_E-4) \) | \( \frac{4}{4+10(V_E-4)} \) | --- ## **Final Answers** ### **1) \( I_S \) vs \( V_E \) Graph:** - From (0,0) to (1,4): slope = 4 - From (1,4) to (4,4): horizontal - From (4,4) upward: slope = 10 ### **2) \( P_L \) vs \( V_E \) Graph:** - From (0,0) to (1,4): parabola \( 4V_E^2 \) - From (1,4) upward: straight line with slope 4 ### **3) Maximum \( \frac{P_L}{P_E} \) and at what \( V_E \):** - **Maximum \( \frac{P_L}{P_E} = 1 \)** - **Occurs for \( 0 < V_E < 4 \)** --- If you need the actual sketches or further clarification, let me know!

Certainly! Let's provide a detailed step-by-step solution with all calculations and explanations for each part. --- ## **Given Data and Circuit Overview** - The circuit includes three characteristic curves: - \( Z_S \): Source impedance with \( I \) vs. \( V \) - \( Z_L \): Load impedance with \( I \) vs. \( V \) - \( Z_D \): Diode or nonlinear element with \( I \) vs. \( V \) - The characteristic curves are provided with specific points: - \( Z_S \): \( I \) from 0 to 5 as \( V \) from 0 to 2 (slope = 2.5) - \( Z_L \): \( I \) from 0 to 4 as \( V \) from 0 to 1, then constant at 4 - \( Z_D \): \( I \) from 0 to 10 as \( V \) from 4 to higher (slope = 10) - The variables: - \( V_E \): Input or supply voltage - \( I_S \): Source current - \( P_L \): Power consumed in load \( Z_L \) - \( P_E \): Power supplied by the source \( E \) --- ## **Part 1: Draw a graph of \( I_S \) versus \( V_E \)** ### Step 1: Understand the characteristic curves - **\( Z_L \):** - \( I_{Z_L} = \frac{V}{Z_L} \) - From the characteristic: - \( I_{Z_L} \) increases linearly from 0 at \( V=0 \) to 4 at \( V=1 \). - For \( V > 1 \), \( I_{Z_L} \) remains constant at 4. - **\( Z_D \):** - For \( V > 4 \), \( I_{Z_D} = 10(V - 4) \). - For \( V \leq 4 \), \( I_{Z_D} = 0 \). ### Step 2: Express \( I_S \) in terms of \( V_E \) Since the elements are in parallel across \( V_E \), the total source current: \[ I_S = I_{Z_L} + I_{Z_D} \] ### Step 3: Determine \( I_S \) in regions based on \( V_E \) - **Region 1: \( 0 \leq V_E < 1 \)** - \( I_{Z_L} = \frac{V_E}{Z_{L}} \) - From the curve, at \( V = 1 \), \( I_{Z_L} = 4 \), so at \( V_E < 1 \), the relation is linear: \[ I_{Z_L} = 4 V_E \] - \( I_{Z_D} = 0 \) since \( V_E < 4 \) - Therefore, \[ I_S = 4 V_E \] - **Region 2: \( 1 \leq V_E < 4 \)** - \( I_{Z_L} = 4 \) (saturated at maximum) - \( I_{Z_D} = 0 \) (since \( V_E < 4 \)) \[ I_S = 4 \] - **Region 3: \( V_E \geq 4 \)** - \( I_{Z_L} = 4 \) (assuming load remains saturated) - \( I_{Z_D} = 10 (V_E - 4) \) \[ I_S = 4 + 10(V_E - 4) = 10 V_E - 36 \] ### **Final \( I_S \) vs \( V_E \) graph:** - From \( 0 \) to \( 1 \): \[ I_S = 4 V_E \] - Slope = 4, points from (0,0) to (1,4) - From \( 1 \) to \( 4 \): \[ I_S = 4 \] - Horizontal line at \( I_S = 4 \) - From \( 4 \) onwards: \[ I_S = 10 V_E - 36 \] - Slope = 10, starting at \( V_E=4 \), \( I_S=4 \) --- ## **Part 2: Power \( P_L \) in \( Z_L \) as a function of \( V_E \)** ### Step 1: Write \( P_L \) \[ P_L = V_E \times I_{Z_L} \] ### Step 2: Express \( P_L \) in each region - **Region 1: \( 0 \leq V_E < 1 \)** \[ I_{Z_L} = 4 V_E \] \[ P_L = V_E \times 4 V_E = 4 V_E^2 \] - **Region 2: \( 1 \leq V_E < 4 \)** \[ I_{Z_L} = 4 \] \[ P_L = V_E \times 4 = 4 V_E \] - **Region 3: \( V_E \geq 4 \)** \[ I_{Z_L} = 4 \] \[ P_L = V_E \times 4 = 4 V_E \] **Note:** For \( V_E \geq 4 \), power in load continues to increase linearly with \( V_E \). ### **Summary:** | \( V_E \) Range | \( P_L \) expression | Graph shape | |-----------------|----------------------|--------------| | \( 0 \leq V_E < 1 \) | \( 4 V_E^2 \) | Parabola from (0,0) to (1,4) | | \( 1 \leq V_E \) | \( 4 V_E \) | Straight line starting at (1,4) | --- ## **Part 3: Find \( V_E \) where \( \frac{P_L}{P_E} \) is maximum** ### Step 1: Express \( P_E \) The power supplied by the source: \[ P_E = V_E \times I_S \] Using \( I_S \) from Part 1: - For \( 0 \leq V_E < 1 \): \[ P_E = V_E \times 4 V_E = 4 V_E^2 \] - For \( 1 \leq V_E < 4 \): \[ P_E = V_E \times 4 = 4 V_E \] - For \( V_E \geq 4 \): \[ P_E = V_E \times (10 V_E -36) = 10 V_E^2 - 36 V_E \] ### Step 2: Calculate \( \frac{P_L}{P_E} \) - **Region 1: \( 0 \leq V_E < 1 \)** \[ P_L = 4 V_E^2, \quad P_E = 4 V_E^2 \] \[ \frac{P_L}{P_E} = \frac{4 V_E^2}{4 V_E^2} = 1 \] - **Region 2: \( 1 \leq V_E < 4 \)** \[ P_L = 4 V_E, \quad P_E = 4 V_E \] \[ \frac{P_L}{P_E} = 1 \] - **Region 3: \( V_E \geq 4 \)** \[ P_L = 4 V_E \] \[ P_E = 10 V_E^2 - 36 V_E \] Therefore, \[ \frac{P_L}{P_E} = \frac{4 V_E}{10 V_E^2 - 36 V_E} = \frac{4}{10 V_E - 36} \] This ratio varies with \( V_E \) when \( V_E \geq 4 \). ### **Step 3: Find maximum of \( \frac{P_L}{P_E} \)** - For \( V_E < 4 \), ratio = 1 (constant). - For \( V_E \geq 4 \), the ratio: \[ R(V_E) = \frac{4}{10 V_E - 36} \] - To find maximum, analyze at \( V_E = 4 \): \[ R(4) = \frac{4}{10 \times 4 - 36} = \frac{4}{40 - 36} = \frac{4}{4} = 1 \] - As \( V_E \to \infty \): \[ R(V_E) \to 0 \] - The ratio \( R(V_E) \) decreases beyond \( V_E=4 \). ### **Conclusion:** - The maximum \( \frac{P_L}{P_E} \) is **equal to 1**. - **It occurs for all \( V_E \) in the ranges:** \[ 0 < V_E < 4 \] - **At \( V_E=4 \), the ratio remains 1.** --- ## **Final Summary:** | Part | Result | Explanation | |--------|--------|--------------| | **Part 1** | \( I_S \) vs \( V_E \): | Piecewise linear: - \( 0 \leq V_E < 1 \): \( I_S = 4V_E \) - \( 1 \leq V_E < 4 \): \( I_S = 4 \) - \( V_E \geq 4 \): \( I_S = 10V_E - 36 \) | | **Part 2** | \( P_L \) vs \( V_E \): | - \( 0 \leq V_E < 1 \): \( P_L = 4 V_E^2 \) - \( V_E \geq 1 \): \( P_L = 4 V_E \) | | **Part 3** | Max \( \frac{P_L}{P_E} \): | Equal to 1 for all \( 0 < V_E < 4 \). At \( V_E=4 \), ratio = 1. It decreases beyond that. | --- **If needed, I can help you draw the graphs or clarify any step further!**