Let's break this down step by step: --- ## **1. Drawing the Circuit**: Let's clarify the circuit from your description: ``` 10V 5V --- --- | | [4Ω] [2Ω] | | +--------+----------+ | (A) / \ [6Ω] [3Ω] | | GND GND ``` - **10 V source** in series with **4 Ω** connects to node A. - **5 V source** in series with **2 Ω** connects to node A. - From **node A**, two branches: **6 Ω** to ground, and **3 Ω** to ground. --- ## **2. Assigning Currents** Let's define: - \( I_1 \): Current from 10 V source through 4 Ω into node A. - \( I_2 \): Current from 5 V source through 2 Ω into node A. - \( I_3 \): Current from node A through 6 Ω to ground. - \( I_4 \): Current from node A through 3 Ω to ground. At node A: \( I_1 + I_2 = I_3 + I_4 \) --- ## **3. Write Two Loop Equations (Branch Current Method)** Let's form two loops: ### **Loop 1: 10V → 4Ω → Node A → 6Ω → Ground → back to 10V** - Start at negative terminal of 10 V source, go through source, 4 Ω, node A, 6 Ω, ground, back to source. \[ -10V + 4I_1 + 6I_3 = 0 \] But \( I_1 \) is the current in 4 Ω, \( I_3 \) is the current down the 6 Ω branch. From KCL at node A, \( I_1 \) goes into node A, \( I_3 \) leaves node A—**but in this loop, only one path is taken at a time**. So let's define mesh currents for clarity: Let: - \( I_a \): Mesh current, 10V source → 4Ω → Node A → 6Ω → Ground → Back to 10V - \( I_b \): Mesh current, 5V source → 2Ω → Node A → 3Ω → Ground → Back to 5V **Note:** The resistors at node A are not shared between loops, so the mesh currents in the two left-side loops do not overlap in the parallel resistors; node A is a junction. So, let's use **node voltage (nodal analysis)** as it's more straightforward. But since you asked for both, let's continue. ### **Loop 1 (10V Loop):** \[ -10 + 4I_1 + 6I_3 = 0 \] ### **Loop 2 (5V Loop):** \[ -5 + 2I_2 + 3I_4 = 0 \] But from KCL at node A: \[ I_1 + I_2 = I_3 + I_4 \] But we have 4 unknowns and 3 equations. However, the current through each branch can be linked via node A voltage. --- ## **4. Using Branch Current Method with Node A Voltage** Let \( V_A \) be the voltage at node A with respect to ground. - Current through 4Ω: \( I_1 = \frac{10 - V_A}{4} \) - Current through 2Ω: \( I_2 = \frac{5 - V_A}{2} \) - Current through 6Ω: \( I_3 = \frac{V_A - 0}{6} \) - Current through 3Ω: \( I_4 = \frac{V_A - 0}{3} \) KCL at node A: \[ I_1 + I_2 = I_3 + I_4 \] Insert the above expressions: \[ \frac{10 - V_A}{4} + \frac{5 - V_A}{2} = \frac{V_A}{6} + \frac{V_A}{3} \] --- ## **5. Solve for \( V_A \) (Nodal Analysis)** Combine the right side: \[ \frac{V_A}{6} + \frac{V_A}{3} = \frac{V_A}{6} + \frac{2V_A}{6} = \frac{3V_A}{6} = \frac{V_A}{2} \] Now, left side: \[ \frac{10 - V_A}{4} + \frac{5 - V_A}{2} = \frac{10 - V_A}{4} + \frac{10 - 2V_A}{4} = \frac{10 - V_A + 10 - 2V_A}{4} = \frac{20 - 3V_A}{4} \] But that's not correct, let's be more careful: \[ \frac{10 - V_A}{4} + \frac{5 - V_A}{2} \] To combine, convert to common denominator (4): \[ \frac{10 - V_A}{4} + \frac{2(5 - V_A)}{4} = \frac{10 - V_A + 10 - 2V_A}{4} = \frac{20 - 3V_A}{4} \] So the equation: \[ \frac{20 - 3V_A}{4} = \frac{V_A}{2} \] Multiply both sides by 4: \[ 20 - 3V_A = 2V_A \] \[ 20 = 5V_A \] \[ V_A = 4\,\text{V} \] --- ## **6. Find Current Through 3Ω Resistor** \[ I_4 = \frac{V_A}{3} = \frac{4}{3} = 1.\underline{3}3\,\text{A} \] --- ## **7. Final Answers** - **Node A voltage:** \( \boxed{4\,\text{V}} \) - **Current through 3Ω resistor:** \( \boxed{1.33\,\text{A}} \) (to 2 decimal places) --- ## **8. Brief Explanation: Why Nodal Analysis is More Efficient** **Nodal analysis** is more efficient here because: - The circuit contains multiple sources and resistors meeting at a single node (node A). - Nodal analysis allows you to write a single equation for node A using KCL, directly relating all source and resistor currents through the node voltage. - The branch current (loop/mesh) method would require defining and solving multiple loop currents and more simultaneous equations, especially with sources not in the same mesh. - In this circuit, nodal analysis quickly gives the answer with only one main equation. --- ## **Summary Table** | Quantity | Value | |----------------------------|--------------| | Node A voltage \((V_A)\) | \(4\,\text{V}\) | | Current through \(3\,\Omega\) | \(1.33\,\text{A}\) | --- ### **Loop Equations (as requested):** 1. \( -10 + 4I_1 + 6I_3 = 0 \) 2. \( -5 + 2I_2 + 3I_4 = 0 \) But using nodal analysis is the most direct approach here. --- **If you need the current directions or other branch currents, just ask!**