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We are tasked with determining the current \( I_2 \) in the circuit, using the provided nodal voltage \( V_A = 12 \, V \), the source voltages \( E_1 = 12 \, V \) and \( E_2 = 13 \, V \), and the resistances \( R_1 = 38 \, \Omega \), \( R_2 = 12 \, \Omega \), and \( R_3 = 26 \, \Omega \).

To solve for \( I_2 \), we can apply **Kirchhoff's Current Law (KCL)** at node \( V_A \) and Ohm's law for the branches.

### Step-by-Step Approach:

1. **Current through \( R_1 \):**
   \[
   I_1 = \frac{E_1 - V_A}{R_1} = \frac{12 \, V - 12 \, V}{38 \, \Omega} = 0 \, A
   \]

2. **Current through \( R_3 \) (from \( V_A \) to ground):**
   \[
   I_3 = \frac{V_A}{R_3} = \frac{12 \, V}{26 \, \Omega} = 0.4615 \, A
   \]

3. **Current through \( R_2 \) (which is \( I_2 \)):**
   Applying KCL at node \( V_A \), the sum of the currents leaving the node equals zero:
   \[
   I_1 + I_2 + I_3 = 0
   \]
   Since \( I_1 = 0 \), the equation simplifies to:
   \[
   I_2 = -I_3 = -0.4615 \, A
   \]
   This negative sign indicates the current direction \( I_2 \) is opposite to the assumed direction in the diagram.

Therefore, the current \( I_2 \) is approximately \( -0.4615 \, A \), or \( 461.5 \, mA \) in the opposite direction of what is assumed in the diagram.

Would you like more details or have any questions?

Here are 5 relative questions that expand upon this topic:
1. How does Kirchhoff's Current Law apply to complex circuits?
2. What would change in the current calculation if \( E_1 \) or \( E_2 \) were altered?
3. How do you calculate the power dissipated in \( R_2 \)?
4. How does Ohm's Law help in analyzing multi-loop circuits?
5. What is the significance of a negative current in circuit analysis?

**Tip:** Always double-check assumed current directions when solving with KCL, as negative values indicate a direction opposite to what was assumed.

For the circuit below, determine the current I2 if the nodal voltage VA = 12 V, E2 = 13 V, R1 = 38 Ω, R2 = 12 Ω, and R3 = 26 Ω.

Learn to determine the current I2 in a circuit with a given nodal voltage VA = 12 V, source voltages E1 = 12 V and E2 = 13 V, and resistances R1 = 38 Ω, R2 = 12 Ω, and R3 = 26 Ω. This problem covers circuit analysis, Kirchhoff's Current Law, and Ohm's Law, making it ideal for electrical engineering students.

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