Let's solve the problem step by step: ### **Step 1: Simplify the Circuit** We'll combine resistors step by step, starting from the right side of the circuit. #### **A. Combine the two 12 Ω resistors in parallel:** \[ R_{12p} = \frac{12 \times 12}{12 + 12} = \frac{144}{24} = 6 \, \Omega \] #### **B. Add the 6 Ω resistor in series with the result:** \[ R_{B} = 6 + 6 = 12 \, \Omega \] #### **C. Combine the 8 Ω resistor in parallel with the result:** \[ R_{C} = \frac{8 \times 12}{8 + 12} = \frac{96}{20} = 4.8 \, \Omega \] #### **D. Add the 5 Ω resistor in series with the result:** \[ R_{D} = 5 + 4.8 = 9.8 \, \Omega \] #### **E. Combine the 4 Ω and 6 Ω resistors in parallel:** \[ R_{4p6} = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4 \, \Omega \] #### **F. Add the 3 Ω resistor in series with the result:** \[ R_{E} = 3 + 2.4 = 5.4 \, \Omega \] #### **G. Now, combine the results from steps D and E in parallel:** \[ R_{eq} = \frac{5.4 \times 9.8}{5.4 + 9.8} = \frac{52.92}{15.2} \approx 3.48 \, \Omega \] --- ### **Step 2: Find the Power Supplied** \[ P = \frac{V^2}{R_{eq}} \] Given: \( V = 25 \) V \[ P = \frac{25^2}{3.48} \approx \frac{625}{3.48} \approx 179.6 \, \text{W} \] --- ## **Final Answers** - **Equivalent Resistance (\( R_{eq} \))**: **3.48 Ω** - **Power Supplied by Source**: **179.6 W** If you need step-by-step drawings or further details, let me know!

Let's walk through the problem together in a clear and straightforward way. **Step 1: Simplify the Circuit** Our goal is to find the equivalent resistance \( R_{eq} \) seen from terminals (a, b), and then determine how much power the voltage source supplies. 1. **Identify the resistors and their connections.** The circuit has several resistors connected in various combinations—some in series, some in parallel. 2. **Combine resistors step-by-step:** - **Parallel of two 12 Ω resistors:** \[ R_{1} = \frac{12 \times 12}{12 + 12} = \frac{144}{24} = 6\, \Omega \] - **Add the 6 Ω resistor in series:** \[ R_{2} = 6 + 6 = 12\, \Omega \] - **Parallel of the 8 Ω resistor with this 12 Ω:** \[ R_{3} = \frac{8 \times 12}{8 + 12} = \frac{96}{20} = 4.8\, \Omega \] - **Add the 5 Ω resistor in series:** \[ R_{4} = 5 + 4.8 = 9.8\, \Omega \] - **Parallel of 4 Ω and 6 Ω resistors:** \[ R_{5} = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\, \Omega \] - **Add the 3 Ω resistor in series:** \[ R_{6} = 3 + 2.4 = 5.4\, \Omega \] 3. **Combine the two results (from steps involving \( R_4 \) and \( R_5 \)):** These are in parallel: \[ R_{total} = \frac{R_{4} \times R_{6}}{R_{4} + R_{6}} = \frac{9.8 \times 5.4}{9.8 + 5.4} \approx \frac{52.92}{15.2} \approx 3.48\, \Omega \] **So, the equivalent resistance \( R_{eq} \) seen from the terminals is approximately 3.48 Ω.** --- **Step 2: Find the Power Supplied** Given the voltage source is 25 V: \[ P = \frac{V^2}{R_{eq}} = \frac{25^2}{3.48} \approx \frac{625}{3.48} \approx 179.6\, \text{W} \] **Final result:** - **Equivalent Resistance:** approximately **3.48 Ω** - **Power supplied by the voltage source:** approximately **180 W** Let me know if you'd like a detailed diagram or further explanation!