Let's solve the circuit step-by-step: ### **Step 1: Understanding the Circuit** - The circuit contains a battery (9 V), five resistors (R1 to R5), and a capacitor (C1). - Since the circuit is DC and the capacitor is in steady state (fully charged), it acts as an **open circuit** (no current through C1 branch). - So, we can **ignore C1** for DC steady-state analysis. --- ### **Step 2: Redraw the Circuit Without C1** The circuit reduces to: ``` ---R1--R2---- | | |--R3--R4----| | | |----R5------| | | | | +----[ V1 ]--+ ``` --- ### **Step 3: Simplify the Circuit** #### **a. Combine R3 and R4 (Series):** - \( R_{34} = R3 + R4 = 300\,\Omega + 400\,\Omega = 700\,\Omega \) #### **b. Combine R2 and R34 (Parallel):** - \( \frac{1}{R_{234}} = \frac{1}{R2} + \frac{1}{R_{34}} = \frac{1}{200} + \frac{1}{700} \) - \( \frac{1}{R_{234}} = \frac{700 + 200}{140000} = \frac{900}{140000} \) - \( R_{234} = \frac{140000}{900} \approx 155.56\,\Omega \) #### **c. Now, R1, R234, and R5 are in series:** - \( R_{eq} = R1 + R_{234} + R5 = 100 + 155.56 + 500 = 755.56\,\Omega \) --- ### **Step 4: Find the Total Current** - \( I_{total} = \frac{V}{R_{eq}} = \frac{9}{755.56} \approx 0.0119\,\text{A} = 11.9\,\text{mA} \) --- ### **Step 5: Find Voltage Drops** #### **a. Across R1:** - \( V_{R1} = I_{total} \times R1 = 0.0119 \times 100 = 1.19\,\text{V} \) #### **b. Across R5:** - \( V_{R5} = I_{total} \times R5 = 0.0119 \times 500 = 5.95\,\text{V} \) #### **c. Across R234:** - \( V_{R234} = I_{total} \times R_{234} = 0.0119 \times 155.56 = 1.85\,\text{V} \) --- ### **Step 6: Find Currents Through Each Resistor** #### **a. R1 and R5 (Series):** - \( I_{R1} = I_{R5} = I_{total} = 0.0119\,\text{A} \) #### **b. In the Parallel Section (R2 and R34):** - Voltage across both = \( V_{R234} = 1.85\,\text{V} \) ##### **Current through R2:** - \( I_{R2} = \frac{1.85}{200} = 0.00925\,\text{A} \) ##### **Current through R34:** - \( I_{R34} = \frac{1.85}{700} = 0.00264\,\text{A} \) ##### **Current through R3 and R4 (Series, same current):** - \( I_{R3} = I_{R4} = I_{R34} = 0.00264\,\text{A} \) --- ### **Step 7: Final Answers (Rounded to 2 Decimal Places)** | Resistor | Current (A) | Current (mA) | |----------|-------------|--------------| | R1 | 0.0119 | 11.9 | | R2 | 0.00925 | 9.25 | | R3 | 0.00264 | 2.64 | | R4 | 0.00264 | 2.64 | | R5 | 0.0119 | 11.9 | --- ## **Summary Table** - **R1:** 0.0119 A (11.9 mA) - **R2:** 0.00925 A (9.25 mA) - **R3:** 0.00264 A (2.64 mA) - **R4:** 0.00264 A (2.64 mA) - **R5:** 0.0119 A (11.9 mA) --- **Explanation:** - Capacitor is ignored for DC steady-state. - Combine resistors step by step (series, then parallel, then series again). - Use Ohm’s law and rules for parallel/series circuits to find total and individual currents. **If you need the voltage across any resistor, just multiply its current by its resistance.**