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To solve for \( V_a \) in this circuit using Kirchhoff's Current Law (KCL), we first analyze the circuit layout and assign currents to each branch. Here's the step-by-step solution:

### Step 1: Labeling Currents
Let's denote:
- \( I_1 \): The current flowing from the 8V source, through the 125Ω resistor.
- \( I_2 \): The current flowing from the 12V source through the 500Ω resistor connected to \( V_a \).
- \( I_3 \): The current flowing through the 250Ω resistor between the two branches.

### Step 2: Set Up Node Equations at \( V_a \)
By KCL, the sum of currents entering a node equals the sum of currents leaving the node. We choose the node at \( V_a \) and write the KCL equation.

At \( V_a \), the currents flowing into the node should equal the currents flowing out. We have:
\[
I_1 = I_2 + I_3
\]

### Step 3: Express Currents in Terms of Voltages
Using Ohm's law \( V = IR \), we express each current in terms of \( V_a \):

1. **Current \( I_1 \):**
   \[
   I_1 = \frac{8V - V_a}{125 \, \Omega}
   \]

2. **Current \( I_2 \):**
   \[
   I_2 = \frac{12V - V_a}{500 \, \Omega}
   \]

3. **Current \( I_3 \):**
   \[
   I_3 = \frac{V_a}{250 \, \Omega}
   \]

### Step 4: Substitute Currents into the KCL Equation
Substitute the expressions for \( I_1 \), \( I_2 \), and \( I_3 \) into the KCL equation:

\[
\frac{8 - V_a}{125} = \frac{12 - V_a}{500} + \frac{V_a}{250}
\]

### Step 5: Clear the Denominator
To simplify, multiply through by the least common multiple of the denominators, which is 500:

\[
4(8 - V_a) = (12 - V_a) + 2V_a
\]

Expanding and simplifying:

\[
32 - 4V_a = 12 - V_a + 2V_a
\]

Combine terms involving \( V_a \):

\[
32 - 4V_a = 12 + V_a
\]

Add \( 4V_a \) to both sides:

\[
32 = 12 + 5V_a
\]

Subtract 12 from both sides:

\[
20 = 5V_a
\]

Divide by 5:

\[
V_a = 4V
\]

### Solution
The voltage \( V_a \) is **4 volts**.

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Would you like further details or have any questions on specific steps?

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Here are 5 related questions to deepen your understanding:
1. How does KCL help in solving complex circuits with multiple sources?
2. Why is it necessary to express all currents in terms of voltage when applying KCL?
3. How would the solution change if one of the voltage sources had a different value?
4. How can Kirchhoff's Voltage Law (KVL) be used in conjunction with KCL in circuit analysis?
5. What are the effects of resistor values on the voltage at node \( V_a \)?

**Tip:** When solving circuits with multiple sources, consider solving simpler sub-circuits separately, if possible, and then combining results.

Refer the electrical circuit shown below. Find Va using KCL.

Solve for the voltage Va in a circuit using Kirchhoff's Current Law (KCL). This example covers the application of Ohm's Law and KCL to analyze currents and voltages in a circuit with multiple resistors and voltage sources.

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