DC circuits and network theorems notes — Unit 1

An electrical circuit is a closed path through which direct current (DC) flows. It consists of active elements (which supply energy, like batteries) and passive elements (which consume or store energy, like resistors, capacitors, inductors).

⇒Basic Terminologies

• Node: A junction where two or more circuit elements are connected.
• Branch: A part of the circuit that lies between two nodes.
• Loop: Any closed path in a circuit.
• Mesh: A closed path that does not contain any other closed path inside it.

Ohm's law states that at a constant temperature, the current ($I$) flowing through a conductor is directly proportional to the potential difference ($V$) across its ends.

V ∝ I  ⟹  V = IR

Where R is Resistance (measured in Ohms, Ω)

⇒Limitations of Ohm's Law

Ohm's law is NOT a universal law. It fails in several scenarios:

• Non-linear devices: Diodes, transistors, and thyristors do not have a linear V-I relationship.
• Non-metallic conductors: Electrolytes and gases.
• Temperature variation: If the temperature changes, resistance changes, and the V-I graph is no longer a straight line (e.g., an incandescent bulb filament).

Resistance depends on the physical dimensions of the conductor: $R = \rho \frac{L}{A}$, where $\rho$ is resistivity.

⇒Temperature Coefficient of Resistance ($\alpha$)

For metals, resistance increases with temperature. The variation is given by:

R_t = R₀ [1 + α₀(T - T₀)]

Where:
R_t = Resistance at temp T
R₀ = Resistance at reference temp T₀
α₀ = Temperature coefficient at T₀ (measured in /°C)

⇒Series Combination

Resistors are connected end-to-end. The same current flows through all resistors, but voltage divides.

R_eq = R₁ + R₂ + R₃ + ...

⇒Parallel Combination

Resistors are connected across the same two nodes. The same voltage appears across all resistors, but current divides.

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + ...
For two resistors: R_eq = (R₁R₂) / (R₁ + R₂)

⇒Voltage Divider Rule (VDR) for Series

If total voltage $V$ is applied across $R_1$ and $R_2$ in series, the voltage across $R_1$ is:

V₁ = V * [ R₁ / (R₁ + R₂) ]

⇒Current Divider Rule (CDR) for Parallel

If total current $I$ enters a parallel combination of $R_1$ and $R_2$, the current through $R_1$ is:

I₁ = I * [ R₂ / (R₁ + R₂) ]

KCL (Kirchhoff's First Law) states that the algebraic sum of all currents meeting at a node (junction) in an electrical circuit is zero.

Σ I = 0
Or: Sum of incoming currents = Sum of outgoing currents

⇒Physical Principle

KCL is based on the Law of Conservation of Charge. A node cannot store charge; whatever charge flows in per second must flow out.

KVL (Kirchhoff's Second Law) states that in any closed loop of an electrical circuit, the algebraic sum of all voltages (EMFs and voltage drops) is zero.

Σ V = 0
Σ EMFs + Σ IR drops = 0

⇒Physical Principle

KVL is based on the Law of Conservation of Energy. The total energy supplied by sources in a closed loop equals the total energy consumed by the elements in that loop.

To solve circuits using KVL, you must stick to a strict sign convention. The easiest method is tracking "potential rises" as positive and "potential drops" as negative as you walk around the loop.

• Batteries: Going from negative (-) to positive (+) terminal is a RISE (+V). Going from + to - is a DROP (-V).
• Resistors: Going in the direction of assumed current is a DROP (-IR). Going against the assumed current is a RISE (+IR).

In Mesh Analysis, we assign a circulating current (e.g., $i_1, i_2$ clockwise) to every mesh and write KVL equations for each.

Nodal Analysis uses KCL to find the node voltages in a circuit. It's often faster than Mesh Analysis when there are many parallel branches.

⇒Steps:

• 1. Identify all nodes in the circuit.
• 2. Select one node as the Reference Node (Ground) and assign it 0V.
• 3. Assign voltage variables ($V_1, V_2$) to the remaining nodes.
• 4. Assume all currents are leaving the node (unless a current source dictates otherwise).
• 5. Write KCL equations using Ohm's Law: $I = (V_{node} - V_{adjacent}) / R$.
• 6. Solve the system of linear equations.

Often, resistors are connected in a bridge-like configuration that is neither strictly series nor parallel. We convert between Star (Y) and Delta (Δ) networks to simplify them.

⇒Star to Delta (Y ➝ Δ)

Given a Star network with $R_A, R_B, R_C$, the equivalent Delta resistors ($R_{AB}, R_{BC}, R_{CA}$) are:

R_AB = R_A + R_B + (R_A·R_B) / R_C
R_BC = R_B + R_C + (R_B·R_C) / R_A
R_CA = R_C + R_A + (R_C·R_A) / R_B

⇒Delta to Star (Δ ➝ Y)

Given a Delta network with $R_{AB}, R_{BC}, R_{CA}$, the equivalent Star resistors ($R_A, R_B, R_C$) are:

R_A = (R_AB · R_CA) / (R_AB + R_BC + R_CA)
R_B = (R_AB · R_BC) / (R_AB + R_BC + R_CA)
R_C = (R_BC · R_CA) / (R_AB + R_BC + R_CA)

While Kirchhoff's laws can solve any circuit, the math becomes extremely tedious (solving $4\times4$ matrices for 4 meshes). Network theorems provide shortcuts to analyze complex circuits.

The "Big Four" Theorems you must master for exams:

• Superposition Theorem: Analyzes circuits with multiple sources.
• Thevenin's Theorem: Replaces a complex circuit with a single voltage source and a series resistor.
• Norton's Theorem: Replaces a complex circuit with a single current source and a parallel resistor.
• Maximum Power Transfer: Determines the load required to extract maximum power.

Statement: In any linear, bilateral network containing two or more independent sources, the response (current or voltage) in any branch is equal to the algebraic sum of the responses caused by each independent source acting alone.

⇒How to apply:

• 1. Select one independent source and turn off all others.
• 2. Turn off voltage sources by short-circuiting them (V=0).
• 3. Turn off current sources by open-circuiting them (I=0).
• 4. Calculate the current/voltage in the desired branch.
• 5. Repeat for all sources.
• 6. Algebraically add the results (respecting current directions).

While superposition is incredibly useful for finding voltages and currents, there is a massive trap students fall into regarding Power.

Statement: Any linear, bilateral two-terminal network can be replaced by an equivalent circuit consisting of a single voltage source ($V_{th}$) in series with a single resistance ($R_{th}$).

⇒Steps to find Thevenin Equivalent:

• 1. Remove the load resistor ($R_L$) from the circuit terminals A and B.
• 2. Find $V_{th}$: Calculate the open-circuit voltage across terminals A and B. This is $V_{th}$.
• 3. Find $R_{th}$: Turn off all independent sources (Short V-sources, Open I-sources). Calculate the equivalent resistance looking back into terminals A and B.
• 4. Reconnect $R_L$. The load current is $I_L = V_{th} / (R_{th} + R_L)$.

Finding $V_{th}$ is usually the hardest part for students. Once the load is removed, no current flows through the branch containing the open terminals.

Therefore, there is no voltage drop across resistors that are strictly in series with the open terminals. You must use KVL/KCL on the remaining closed loops to find the node voltages, and then write a KVL equation from terminal A to terminal B to find $V_{AB} = V_{th}$.

Statement: Any linear, bilateral two-terminal network can be replaced by an equivalent circuit consisting of a single current source ($I_N$) in parallel with a single resistance ($R_N$).

⇒Steps to find Norton Equivalent:

• 1. Remove the load resistor ($R_L$) and place a short-circuit wire across terminals A and B.
• 2. Find $I_N$: Calculate the current flowing through this short-circuit wire. This is $I_N$ (Norton current or Short-circuit current).
• 3. Find $R_N$: This is identical to $R_{th}$. Turn off sources and find resistance looking into open terminals A and B.
• 4. Reconnect $R_L$. Using current divider, $I_L = I_N * [R_N / (R_N + R_L)]$.

Statement: A DC voltage source will deliver maximum power to a variable load resistor when the load resistance equals the internal (Thevenin) resistance of the source network ($R_L = R_{th}$).

⇒Mathematical Proof

Power delivered to load: $P = I^2 R_L = [V_{th} / (R_{th} + R_L)]^2 R_L$

To find maximum power, take derivative with respect to $R_L$ and set to zero:

dP/dR_L = [V_{th}² / (R_{th} + R_L)⁴] * [ (R_{th} + R_L)²(1) - R_L(2)(R_{th} + R_L) ] = 0
(R_{th} + R_L) - 2R_L = 0
R_{th} - R_L = 0  ⟹  R_L = R_{th}

When $R_L = R_{th}$, the power delivered is maximum. What is this maximum power value?

P_max = [V_{th} / (R_{th} + R_{th})]² R_{th}
P_max = (V_{th} / 2R_{th})² R_{th}
P_max = V_{th}² / (4 R_{th})

⇒Efficiency at Max Power

At max power transfer, half the power is consumed by the load, and half is wasted as heat inside the source's internal resistance. Therefore, the efficiency is exactly 50%. This is why it is used in low-power communications (like audio amplifiers matching speaker impedance), but NEVER in power grids (where 50% loss would be disastrous).

⇒Unit 1 Mastery

• Can you write KVL equations for a 3-mesh circuit?
• Can you perform Star-to-Delta and Delta-to-Star conversions?
• State the Superposition theorem and explain how to handle power calculations.
• Outline the 4 steps to find a Thevenin Equivalent circuit.
• Prove that maximum power is transferred when $R_L = R_{th}$, and state the efficiency.