## Problem Explanation You are asked to **determine the current through the 1kΩ resistor** in the given AC circuit. Let's solve this step by step using **phasor analysis**. --- ### **Step 1: Identify Circuit Elements** - **Resistors**: 1 kΩ, 4 kΩ, 2 kΩ - **Voltage sources**: - AC Source 1: \( 6 \angle ^\circ \) V - AC Source 2: \( 10 \angle ^\circ \) V (between ground and node between 4kΩ and 2kΩ resistors) --- ### **Step 2: Assign Node Voltages** Let’s label the top left node (after 1kΩ and before 4kΩ resistor) as **Node A** and top right node (after 4kΩ, before 2kΩ) as **Node B**. Ground is at the bottom. --- ### **Step 3: Write Node Voltage Equations (Nodal Analysis)** Let: - \( V_A \): Voltage at Node A - \( V_B \): Voltage at Node B #### **Node A:** - Incoming sources: \( 6 \angle ^\circ \) V (directly connected to left side of 1kΩ resistor) - Outgoing currents: through 1kΩ and 4kΩ Apply KCL at Node A: \[ \frac{V_A - 6}{1k} + \frac{V_A - V_B}{4k} = \] #### **Node B:** - Connected to 2kΩ to ground and receives 10∠° V from a source. As per the circuit, - \( V_B = 10 \angle ^\circ \) V (since the voltage source is between ground and Node B) --- ### **Step 4: Substitute Known Values** From above, - \( V_B = 10 \angle ^\circ \) - \( V_A \): unknown Plug into Node A equation: \[ \frac{V_A - 6}{1k} + \frac{V_A - 10}{4k} = \] Multiply both sides by \( 4k \) to clear denominators: \[ 4(V_A - 6) + (V_A - 10) = \] \[ 4V_A - 24 + V_A - 10 = \] \[ 5V_A = 34 \] \[ V_A = 6.8 \text{ V} \] --- ### **Step 5: Find the Current Through 1kΩ Resistor** The current \( I \) through the 1kΩ resistor is: \[ I = \frac{V_A - 6}{1k} \] Plug in \( V_A = 6.8 \): \[ I = \frac{6.8 - 6}{1k} = \frac{.8}{100} = .0008~\text{A} = .8~\text{mA} \] Expressed as a **phasor** (since all sources are at ° phase): \[ \boxed{I = .8 \angle ^\circ~\text{mA}} \] --- ## **Final Answer** ### **The current through the 1 kΩ resistor is:** \[ \boxed{I = .8 \angle ^\circ~\text{mA}} \] If you need more details or have questions about any step, let me know!