Let's analyze the circuit step by step. ## **Given Values** - \( R_1 = 9.1 \text{k}\Omega \) - \( R_2 = 3 \text{k}\Omega \) - \( R_3 = 4.7 \text{k}\Omega \) - \( R_4 = 1 \text{k}\Omega \) - \( R_5 = 5.1 \text{k}\Omega \) - \( V_1 = 8 \text{V} \) (Battery) --- ## **Step 1: Redraw and Label the Circuit** We’ll label the main nodes as follows: - **Node A**: Top left (between R1, R2, and R4) - **Node B**: Top right (between R1, R3, and R5) - **Node C**: Bottom (ground) --- ## **Step 2: Find Equivalent Resistance** ### **A. Combine R4 and R5 with the rest** R4 and R5 are both connected from nodes A and B respectively to the bottom node (ground). #### **R2 and R4 are in series (left branch):** \[ R_{24} = R_2 + R_4 = 3\text{k}\Omega + 1\text{k}\Omega = 4\text{k}\Omega \] #### **R3 and R5 are in series (right branch):** \[ R_{35} = R_3 + R_5 = 4.7\text{k}\Omega + 5.1\text{k}\Omega = 9.8\text{k}\Omega \] #### **R_{24} and R_{35} are in parallel:** \[ R_{eq1} = \left(\frac{1}{R_{24}} + \frac{1}{R_{35}}\right)^{-1} = \left(\frac{1}{4\text{k}\Omega} + \frac{1}{9.8\text{k}\Omega}\right)^{-1} \] Calculate: - \(\frac{1}{4} = 0.25\) - \(\frac{1}{9.8} \approx 0.102\) - \(0.25 + 0.102 = 0.352\) - \(R_{eq1} = \frac{1}{0.352} \approx 2.84\text{k}\Omega\) #### **Now add R1 in series:** \[ R_{total} = R_1 + R_{eq1} = 9.1\text{k}\Omega + 2.84\text{k}\Omega = 11.94\text{k}\Omega \] --- ## **Step 3: Find Total Current from the Battery** \[ I_{total} = \frac{V_1}{R_{total}} = \frac{8\text{V}}{11.94\text{k}\Omega} \approx 0.670\text{mA} \] --- ## **Step 4: Find Voltage Drops** ### **A. Voltage across R1:** \[ V_{R1} = I_{total} \times R_1 = 0.670\text{mA} \times 9.1\text{k}\Omega \approx 6.10\text{V} \] ### **B. Voltage across the parallel network (i.e., across R2+R4 and R3+R5):** \[ V_{parallel} = V_1 - V_{R1} = 8\text{V} - 6.10\text{V} = 1.90\text{V} \] --- ## **Step 5: Branch Currents** ### **A. Left Branch (R2 + R4):** \[ I_{left} = \frac{V_{parallel}}{R_{24}} = \frac{1.90\text{V}}{4\text{k}\Omega} = 0.475\text{mA} \] ### **B. Right Branch (R3 + R5):** \[ I_{right} = \frac{V_{parallel}}{R_{35}} = \frac{1.90\text{V}}{9.8\text{k}\Omega} = 0.194\text{mA} \] --- ## **Step 6: Check Current Conservation** \[ I_{left} + I_{right} \approx 0.475\text{mA} + 0.194\text{mA} = 0.669\text{mA} \approx I_{total} \] --- ## **Summary Table** | Component | Value | Voltage Drop (V) | Current (mA) | |---------------------|--------------|------------------|--------------| | R1 | 9.1 kΩ | 6.10 | 0.67 | | R2 + R4 (series) | 3 kΩ + 1 kΩ | 1.90 | 0.475 | | R3 + R5 (series) | 4.7 kΩ + 5.1 kΩ | 1.90 | 0.194 | --- ## **Conclusion** - **Total current from the battery:** **0.67 mA** - **Voltage across R1:** **6.10 V** - **Current through R2+R4:** **0.475 mA** - **Current through R3+R5:** **0.194 mA** If you have any specific questions (e.g., voltage at V1 or V2, voltage across a specific resistor), let me know!