Solving a Circuit Using KVL and KCL 

Use Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) to find V_x​ and I_y.

Solving a Circuit Using KVL and KCL 

Applying Kirchhoff’s Voltage Law (KVL)

KVL states: The sum of all voltages in a closed loop is zero.

Loop 1: 20V Source, 2Ω Resistor, and 5A Current Source

We start with Loop 1 which is shown below.

Solve Using KVl and KCL - Loop 1

The KVL equation is:

    \[-20V - (2\Omega \cdot 2A) + V_{5A} = 0\]

Simplify:

    \[-20V - 4V + V_{5A} = 0\]

    \[V_{5A} = 16V\]

Loop 2: Voltage AcrossV_{5A}, 3V Source, and 7Ω Resistor

Next, we analyze Loop 2 shown below:

Solve Using KVl and KCL - Loop 2

The equation is:

    \[-16V + 4V - (6\Omega \cdot I_y) = 0\]

Simplify:

    \[-12V -6\Omega \cdot I_y = 0\]

    \[I_y = -2A\]

KCL: Applying Kirchhoff’s Current Law (KCL)

KCL states: The total current entering a node equals the total current leaving the node.

At the middle node, the currents are:

    \[2A + 5A +(-2A) - I_{3\Omega} = 0\]

Simplify:

    \[I_{3\Omega} = 5A\]

Finding V_x

Finally, we calculate V_x using KVL in Loop 3:

Solve Using KVl and KCL - Loop 3

The equation is:

    \[-16V + (3\Omega \cdot 5A) - V_x = 0\]

Simplify:

    \[-16V + 15V - V_x = 0\]

    \[V_x = -1V\]

Using Python to Solve the Circuit

One powerful way to solve electrical circuits is by using the Lcapy library in Python. Lcapy is an open-source Python package designed for symbolic linear circuit analysis and signal processing.

Below is an example of how to use Lcapy to solve the given circuit and find I_y (the current through V_2) and I_{R_1} (the current through the 2Ω resistor).

Python Code to Solve the Circuit

```python
from lcapy import Circuit
cct = Circuit("""
V1 1 0 20; down,
R1 1 N_A 2; right=1.5, i=I_{R_1},
I1 N_A 0_2 5; down, 
W N_A 5; up=0.7, 
R2 N_A 3 3; right=1.5, 
V_x 0_3 3 -1; up, 
V2 5 4 4; right=2, i=I_y,
R4 0_4 4 6; up,
W 0 0_2; right, 
W 0_2 0_3; right, 
W 0_3 0_4; right, 
""")
print(f"Current through R1 (I_R1): {cct.R1.i}")
print(f"Current through V2 (I_y): {cct.V2.i}")

which prints

Current through R1 (I_R1): 2
Current through V2 (I_y): -2

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