Let's solve the equivalent resistance step-by-step: ### Step 1: **Simplify the Rightmost Parallel Combination** First, notice the **15 Ω** and **60 Ω** resistors are in parallel: \[ \frac{1}{R_{1}} = \frac{1}{15} + \frac{1}{60} \] \[ \frac{1}{R_{1}} = \frac{4 + 1}{60} = \frac{5}{60} \] \[ R_1 = \frac{60}{5} = 12\ \Omega \] --- ### Step 2: **Redraw/Simplify the Middle Branch** Now, the circuit becomes: - Top branch: 10 Ω, 6 Ω - Middle branch: 24 Ω, 9 Ω, 12 Ω (from previous step) - Bottom branch: 5 Ω, 6 Ω Notice 9 Ω and the 12 Ω (result from above) are **in series**: \[ R_2 = 9 + 12 = 21\ \Omega \] --- ### Step 3: **Combine the Middle Branch in Parallel with 24 Ω** Now, **24 Ω** and **21 Ω** are in parallel: \[ \frac{1}{R_3} = \frac{1}{24} + \frac{1}{21} \] \[ \frac{1}{R_3} = \frac{21 + 24}{504} = \frac{45}{504} \] \[ R_3 = \frac{504}{45} = 11.2\ \Omega \] --- ### Step 4: **Combine the Top and Bottom Series Resistors** Top: \( 10\ \Omega + 6\ \Omega = 16\ \Omega \) Bottom: \( 5\ \Omega + 6\ \Omega = 11\ \Omega \) --- ### Step 5: **Combine All Three Parallel Branches** Now, the three branches between the two nodes are: - Top: 16 Ω - Middle: 11.2 Ω - Bottom: 11 Ω All are in parallel: \[ \frac{1}{R_{eq}} = \frac{1}{16} + \frac{1}{11.2} + \frac{1}{11} \] Calculate each: - \( \frac{1}{16} = 0.0625 \) - \( \frac{1}{11.2} ≈ 0.0893 \) - \( \frac{1}{11} ≈ 0.0909 \) \[ \frac{1}{R_{eq}} = 0.0625 + 0.0893 + 0.0909 = 0.2427 \] \[ R_{eq} = \frac{1}{0.2427} ≈ 4.12\ \Omega \] --- ### Step 6: **Check for Misinterpretation** Upon reviewing, the question's answer is **23 Ω**, so let's review the connections: - The **6 Ω** resistor in the middle is NOT in series with 9 Ω, but rather, the 6 Ω on top and 6 Ω on bottom are in series with their respective resistors (10 Ω and 5 Ω). - The 24 Ω and 9 Ω are not in parallel directly; the 24 Ω is in parallel with the rest of the network to the right. Let's proceed more carefully: #### Combine the 15 Ω and 60 Ω as before: \( 12\ \Omega \) #### 9 Ω and 12 Ω are in series: \( 9 + 12 = 21\ \Omega \) #### Now, 24 Ω is in parallel with 21 Ω: \[ \frac{1}{R_A} = \frac{1}{24} + \frac{1}{21} = \frac{21 + 24}{504} = \frac{45}{504} \] \[ R_A = \frac{504}{45} = 11.2\ \Omega \] #### Now, 6 Ω (middle horizontal) is in series with \( R_A \): \( 6 + 11.2 = 17.2\ \Omega \) #### Now, 10 Ω (top left) is in series with that: \( 10 + 17.2 = 27.2\ \Omega \) #### 5 Ω (bottom left) is in series with 6 Ω (bottom right): \( 5 + 6 = 11\ \Omega \) #### Now, the two branches (27.2 Ω and 11 Ω) are in parallel: \[ \frac{1}{R_{eq}} = \frac{1}{27.2} + \frac{1}{11} \] \[ \frac{1}{R_{eq}} = 0.0368 + 0.0909 = 0.1277 \] \[ R_{eq} = \frac{1}{0.1277} ≈ 7.83\ \Omega \] --- #### Let's Re-express the Path: - Top path: 10 Ω, then 6 Ω, then (15 Ω and 60 Ω) parallel, then to the right node. - Middle path: 24 Ω, then 9 Ω, then to the right node. - Bottom path: 5 Ω, then 6 Ω, then to the right node. But from the circuit, the 10 Ω and 24 Ω are not in series; they split at the node. --- ## **Final Step:** ### Let's Redraw and Calculate Step by Step: **1. 15 Ω and 60 Ω in parallel:** \[ R_x = \frac{1}{\frac{1}{15} + \frac{1}{60}} = 12\ \Omega \] **2. 9 Ω and 12 Ω in series:** \[ R_y = 9 + 12 = 21\ \Omega \] **3. 21 Ω and 24 Ω in parallel:** \[ R_z = \frac{1}{\frac{1}{21} + \frac{1}{24}} = \frac{1}{\frac{8 + 7}{168}} = \frac{168}{15} = 11.2\ \Omega \] **4. 6 Ω (middle horizontal) in series with R_z:** \[ R_a = 6 + 11.2 = 17.2\ \Omega \] **5. Now, 10 Ω (top left) and 5 Ω (bottom left) are both in series with their own 6 Ω each.** - Top: \( 10 + 6 = 16 \ \Omega \) - Bottom: \( 5 + 6 = 11 \ \Omega \) **6. Now, all three branches (16 Ω, 17.2 Ω, 11 Ω) are in parallel:** \[ \frac{1}{R_{eq}} = \frac{1}{16} + \frac{1}{17.2} + \frac{1}{11} \] \[ \frac{1}{16} = 0.0625 \] \[ \frac{1}{17.2} \approx 0.0581 \] \[ \frac{1}{11} \approx 0.0909 \] \[ \text{Total: } 0.0625 + 0.0581 + 0.0909 = 0.2115 \] \[ R_{eq} = \frac{1}{0.2115} \approx 4.73 \ \Omega \] --- ### **Conclusion** There seems to be some ambiguity in the circuit connections or the answer key. If the answer is **23 Ω**, the only way this happens is if the circuit is interpreted differently (perhaps 6 Ω and 6 Ω are in series, or a branch is omitted). **Based on standard analysis, the equivalent resistance is approximately \( 4.73 \ \Omega \), not 23 Ω.** If you have further clarification on the circuit connections or if there is a misinterpretation, please let me know!