Problem Design: Understanding Kirchhoff's Current Law (KCL)

Given the circuit in the diagram, use Kirchhoff's Current Law (KCL) to solve for the unknown currents $i_1$, $i_2$, and $i_3$ in the circuit.

Problem Setup:

The current values for the main branches are specified as follows:

• $i_a = 12 \, \text{mA}$ (entering the top node)
• $i_b = 8 \, \text{mA}$ (entering the left node)
• $i_c = 9 \, \text{mA}$ (leaving the bottom node)

Using KCL, we know that the current entering and leaving any node must balance (the sum of currents into a node is zero). The relationships between the currents can be determined by analyzing the three different nodes in the circuit.

Questions:

1. Use KCL at the junction where $i_a$, $i_1$, and $i_b$ meet to set up an equation.
2. Apply KCL at the junction where $i_1$, $i_2$, and $i_3$ meet to set up a second equation.
3. Apply KCL at the junction where $i_2$, $i_3$, and $i_c$ meet to set up a third equation.
4. Solve the system of equations to find $i_1$, $i_2$, and $i_3$.

Solution:

Step 1: Apply Kirchhoff's Current Law at each junction.

At the first node (top junction where $i_a$, $i_b$, and $i_1$ meet): $i_a + i_b = i_1$ Substitute the known values: $12 \, \text{mA} + 8 \, \text{mA} = i_1$ Thus, $i_1 = 20 \, \text{mA}$

Step 2: Apply Kirchhoff's Current Law at the second node (middle junction where $i_1$, $i_2$, and $i_3$ meet): $i_1 = i_2 + i_3$ Substitute $i_1 = 20 \, \text{mA}$: $20 \, \text{mA} = i_2 + i_3 \quad \text{(Equation 1)}$

Step 3: Apply Kirchhoff's Current Law at the third node (bottom junction where $i_2$, $i_3$, and $i_c$ meet): $i_2 + i_c = i_3$ Substitute $i_c = 9 \, \text{mA}$: $i_2 + 9 \, \text{mA} = i_3 \quad \text{(Equation 2)}$

Step 4: Solve the system of equations.

From Equation 2, substitute $i_3 = i_2 + 9 \, \text{mA}$ into Equation 1: $20 \, \text{mA} = i_2 + (i_2 + 9 \, \text{mA})$ Simplify: $20 \, \text{mA} = 2i_2 + 9 \, \text{mA}$ $20 \, \text{mA} - 9 \, \text{mA} = 2i_2$ $11 \, \text{mA} = 2i_2$ $i_2 = \frac{11}{2} \, \text{mA} = 5.5 \, \text{mA}$

Now substitute $i_2 = 5.5 \, \text{mA}$ back into Equation 2: $i_3 = 5.5 \, \text{mA} + 9 \, \text{mA} = 14.5 \, \text{mA}$

Final Answers:

• $i_1 = 20 \, \text{mA}$
• $i_2 = 5.5 \, \text{mA}$
• $i_3 = 14.5 \, \text{mA}$

Do you have any questions, or would you like further details?

Here are 5 related questions to practice:

1. What is Kirchhoff's Voltage Law (KVL), and how does it relate to KCL?
2. How would the current distribution change if $i_a$ were reduced to $10 \, \text{mA}$?
3. Can you derive a general formula for any arbitrary current entering and leaving the nodes using KCL?
4. How do resistors in the branches affect the current values in KCL problems?
5. How would the system of equations change if there were an additional current source in the middle branch?

Tip: When applying Kirchhoff's Laws, always ensure the correct sign convention (current entering vs. current leaving) to avoid errors in your equations.