Part I: Circuit Fundamentals | Chapter 1

Ohm's & Kirchhoff's Laws

The governing equations of all resistive circuits: linear relationships, conservation of charge and energy.

Ohm's Law

Ohm's Law states that the voltage across a resistor is proportional to the current flowing through it. For a resistor with resistance \( R \) (in ohms, Ω):

\[ V = IR \]

\( V \) in volts, \( I \) in amperes, \( R \) in ohms (Ω)

The electrical power dissipated as heat in a resistor is:

\[ P = IV = I^2 R = \frac{V^2}{R} \]

Series and Parallel Resistors

Resistors in series (same current through each) combine as:

\[ R_{\text{series}} = R_1 + R_2 + \cdots + R_n \]

Resistors in parallel (same voltage across each) combine as:

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \]

For the special case of two resistors in parallel: \( R_{\text{parallel}} = \dfrac{R_1 R_2}{R_1 + R_2} \).

Voltage Divider

Two series resistors \( R_1 \) and \( R_2 \) driven by \( V_s \) produce an output at the midpoint:

\[ V_{\text{out}} = V_s \cdot \frac{R_2}{R_1 + R_2} \]

Kirchhoff's Laws

Kirchhoff's laws are direct consequences of conservation of energy (KVL) and conservation of charge (KCL). Together they provide a complete framework for writing circuit equations.

Kirchhoff's Voltage Law (KVL)

The algebraic sum of all voltages around any closed loop is zero:

\[ \sum_{\text{loop}} V_k = 0 \]

Equivalently: the sum of voltage rises equals the sum of voltage drops in any closed loop.

Kirchhoff's Current Law (KCL)

The algebraic sum of all currents entering any node is zero:

\[ \sum_{\text{node}} I_k = 0 \]

Equivalently: the total current entering a node equals the total current leaving it.

Node-Voltage Method

The node-voltage method is the most systematic approach to circuit analysis. Steps:

Choose one node as the reference (ground), assign \( V_0 = 0 \).
Assign unknown voltages \( V_1, V_2, \ldots, V_n \) to remaining nodes.
Apply KCL at each unknown node: sum of currents leaving = 0.
Express each branch current using Ohm's Law: \( I = (V_i - V_j)/R \).
Solve the resulting linear system \( \mathbf{Y}\mathbf{V} = \mathbf{I}_s \).

Mesh Analysis

Mesh analysis assigns a mesh current \( I_m \) to each independent loop and writes KVL for each mesh. For a mesh with resistors \( R_{11} \) (self) and shared \( R_{12} \) (mutual):

\[ R_{11} I_1 - R_{12} I_2 = V_{s1} \]

The result is the mesh resistance matrix equation \( \mathbf{R}\mathbf{I} = \mathbf{V}_s \), dual to the node admittance system.

Python: Resistor Network Analysis

Solve a multi-node resistive network using the node-voltage method (NumPy linear algebra), plot V-I curves for several resistor values, and visualise the voltage divider output as a function of \( R_{\text{top}} \).

Node-Voltage Method, V-I Curves & Voltage Divider

Python

resistor_network.py117 lines

import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt

# ---- Node-Voltage Method: solve a resistive ladder network ----
# Circuit: V_s = 12V source at node 1 (ground = node 0)
# R1=100 ohm between node 1 and 2
# R2=200 ohm between node 2 and 3
# R3=150 ohm between node 3 and ground
# R4=300 ohm between node 2 and ground
# R5=220 ohm between node 3 and ground (parallel with R3)

R1, R2, R3, R4, R5 = 100, 200, 150, 300, 220
Vs = 12.0  # volts

# Conductances
G1 = 1/R1; G2 = 1/R2; G3 = 1/R3; G4 = 1/R4; G5 = 1/R5

# Node admittance matrix (nodes 1, 2, 3; node 0 = ground)
# Node 1 is driven by voltage source => V1 = Vs (modify row 1)
# Node 2: (G1+G2+G4)*V2 - G1*V1 - G2*V3 = 0
# Node 3: (G2+G3+G5)*V3 - G2*V2 = 0

Y = np.array([
    [1,       0,            0        ],   # V1 = Vs (source node)
    [-G1,     G1+G2+G4,    -G2      ],   # KCL at node 2
    [0,       -G2,          G2+G3+G5],   # KCL at node 3
])
b = np.array([Vs, 0.0, 0.0])

V = np.linalg.solve(Y, b)
V1, V2, V3 = V

I_R1 = (V1 - V2) / R1
I_R2 = (V2 - V3) / R2
I_R4 = V2 / R4
I_R3 = V3 / R3
I_R5 = V3 / R5
I_total = I_R1

print('=== Node-Voltage Solution ===')
print(f'V1 = {V1:.4f} V (source)')
print(f'V2 = {V2:.4f} V')
print(f'V3 = {V3:.4f} V')
print(f'I_R1 = {I_R1*1000:.3f} mA')
print(f'I_R2 = {I_R2*1000:.3f} mA')
print(f'I_R4 = {I_R4*1000:.3f} mA  (shunt at node 2)')
print(f'I_R3 = {I_R3*1000:.3f} mA  (shunt at node 3)')
print(f'I_R5 = {I_R5*1000:.3f} mA  (shunt at node 3)')
print(f'KCL check at node 2: {(I_R1 - I_R2 - I_R4)*1e6:.2f} uA (should be ~0)')
print(f'KCL check at node 3: {(I_R2 - I_R3 - I_R5)*1e6:.2f} uA (should be ~0)')

# ---- Voltage Divider ----
R_top = np.array([100, 470, 1000, 4700, 10000])
R_bot = 1000.0
V_div = Vs * R_bot / (R_top + R_bot)
print()
print('=== Voltage Divider (R_bot=1k, Vs=12V) ===')
for rt, vd in zip(R_top, V_div):
    print(f'  R_top={rt:5d} ohm => V_out = {vd:.3f} V')

# ---- Plot 1: V-I curves for different resistors ----
V_arr = np.linspace(0, 12, 300)
fig, axes = plt.subplots(1, 3, figsize=(15, 5))
fig.patch.set_facecolor('#0a0a0a')
for ax in axes:
    ax.set_facecolor('#111827')
    for spine in ax.spines.values():
        spine.set_color('#374151')

resistors = [100, 470, 1000, 2200, 4700]
colors = ['#34d399', '#6ee7b7', '#a7f3d0', '#d1fae5', '#ecfdf5']
ax = axes[0]
for R, c in zip(resistors, colors):
    I = V_arr / R * 1000  # mA
    ax.plot(V_arr, I, color=c, lw=2, label=f'{R} Ω')
ax.set_xlabel('Voltage (V)', color='#9ca3af')
ax.set_ylabel('Current (mA)', color='#9ca3af')
ax.set_title('V-I Characteristics', color='#d1d5db', fontsize=12, pad=10)
ax.tick_params(colors='#9ca3af')
ax.legend(fontsize=8, facecolor='#1f2937', edgecolor='#374151', labelcolor='#d1d5db')
ax.grid(True, color='#1f2937', alpha=0.8)

# ---- Plot 2: Voltage divider output vs R_top ----
ax = axes[1]
R_top_range = np.logspace(1, 5, 500)
V_out_range = Vs * R_bot / (R_top_range + R_bot)
ax.semilogx(R_top_range, V_out_range, color='#34d399', lw=2.5)
ax.axhline(Vs/2, color='#f59e0b', ls='--', lw=1.2, label='Vs/2 = 6 V')
ax.set_xlabel('R_top (Ω)', color='#9ca3af')
ax.set_ylabel('V_out (V)', color='#9ca3af')
ax.set_title('Voltage Divider (R_bot=1kΩ)', color='#d1d5db', fontsize=12, pad=10)
ax.tick_params(colors='#9ca3af')
ax.legend(fontsize=8, facecolor='#1f2937', edgecolor='#374151', labelcolor='#d1d5db')
ax.grid(True, color='#1f2937', alpha=0.8)

# ---- Plot 3: Node voltages as bar chart ----
ax = axes[2]
node_labels = ['V1 (source)', 'V2', 'V3']
node_vals = [V1, V2, V3]
bar_colors = ['#10b981', '#34d399', '#6ee7b7']
bars = ax.bar(node_labels, node_vals, color=bar_colors, edgecolor='#374151', linewidth=1.2)
for bar, val in zip(bars, node_vals):
    ax.text(bar.get_x() + bar.get_width()/2, val + 0.15, f'{val:.2f} V',
            ha='center', va='bottom', color='#d1d5db', fontsize=10)
ax.set_ylabel('Voltage (V)', color='#9ca3af')
ax.set_title('Solved Node Voltages', color='#d1d5db', fontsize=12, pad=10)
ax.tick_params(colors='#9ca3af')
ax.set_ylim(0, Vs * 1.15)
ax.grid(True, axis='y', color='#1f2937', alpha=0.8)

plt.tight_layout(pad=2.0)
plt.savefig('output.png', dpi=140, bbox_inches='tight', facecolor='#0a0a0a')
print()
print('Plot saved to output.png')

Click Run to execute the Python code

Code will be executed with Python 3 on the server

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