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The image shows a circuit with currents \( I_1 \), \( I_2 \), and \( I_3 \) meeting at a node. The information provided includes:

- \( A = 5 \, \text{mA} \)
- \( B = 1.8 \, \text{mA} \)
- \( \varphi = 9 \, \text{rad} \)
- \( \theta = 1.9 \, \text{rad} \)

The currents are expressed as:
- \( I_1 = A e^{j\varphi} \)
- \( I_2 = B e^{j\theta} \)

According to Kirchhoff’s Current Law (KCL), the sum of currents entering and leaving a node must be zero:
\[
I_1 + I_2 + I_3 = 0
\]
This implies that:
\[
I_3 = -(I_1 + I_2)
\]

We will calculate the value of \( I_3 \), its magnitude in mA, and provide a solution. Let's proceed with this calculation.The magnitude of \( I_3 \) is approximately \( 6.37 \, \text{mA} \).

Would you like more details or have any other questions?

Here are 5 related questions for further exploration:
1. How is the phase of \( I_3 \) related to the phases of \( I_1 \) and \( I_2 \)?
2. What happens to \( I_3 \) if \( A \) or \( B \) are changed?
3. How would you represent the currents \( I_1 \), \( I_2 \), and \( I_3 \) in polar form?
4. How is Kirchhoff's Current Law (KCL) applied in a more complex circuit?
5. How does the choice of reference direction affect the sign of the currents?

**Tip:** To solve AC circuit problems involving complex numbers, always work in either rectangular or polar form for consistency, and convert between them when needed.

En el nudo de la figura se tiene que A = 5 mA, B = 1.8 mA, φ = 9 rad y θ = 1.9 rad. Calcular entonces el módulo de I3 en mA.

Learn how to calculate the magnitude of the current I3 at a node using Kirchhoff’s Current Law (KCL). This example involves complex numbers, phasors, and basic electrical engineering concepts with given values for A, B, φ, and θ. Suitable for university-level students.

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