Chapter 1 • 7 Hours • Expected Marks: 7M Appears in Q1 (6M) & Q2 (5M) in All Exams

Transients in Electric Circuit

Foundations of network analysis with dependent sources and evaluation of switching initial conditions in dynamic R-L-C circuits.

1.1 Characteristics of Various Network Elements

Concept & Classification

Electric circuits are composed of interconnected elements that store, dissipate, or deliver energy. Understanding their mathematical and physical classifications is essential before setting up system equations.

Classification Definition & Criterion Examples
Linear vs. Non-linear Satisfies the principles of superposition and homogeneity: \(f(a x_1 + b x_2) = a f(x_1) + b f(x_2)\). V-I curve is a straight line through the origin. Linear: Pure resistors, ideal inductors, ideal capacitors.
Non-linear: Diodes, transistors, saturable magnetic cores.
Bilateral vs. Unilateral V-I characteristics are independent of current direction. Conduction behavior is identical in both directions. Bilateral: Resistors, inductors, capacitors.
Unilateral: Diodes, thyristors, operational amplifiers.
Active vs. Passive An active element can deliver net average energy to the external circuit over an infinite time interval (\(\int_{-\infty}^t p(\tau)d\tau \ge 0\) is violated). Active: Voltage/current sources, op-amps, BJT/FET.
Passive: Resistors (absorb energy), Inductors/Capacitors (store energy).
Lumped vs. Distributed Physical dimensions are much smaller than electrical signal wavelength (\(d \ll \lambda\)). KCL/KVL apply directly. Lumped: Standard discrete R, L, C components.
Distributed: Long transmission lines, waveguides.

Energy Storage Element Equations

\[\text{Inductor: } v_L(t) = L \frac{di_L}{dt}, \quad w_L(t) = \frac{1}{2} L i_L^2(t)\] \[\text{Capacitor: } i_C(t) = C \frac{dv_C}{dt}, \quad w_C(t) = \frac{1}{2} C v_C^2(t)\]
Physics of Invariance: Because energy cannot change instantaneously (\(\Delta w = \infty\) requires infinite power \(P = dw/dt\)), inductor current cannot change instantaneously and capacitor voltage cannot change instantaneously.

1.2 & 1.3 Nodal and Mesh Analysis with Dependent Sources

Core Principles

Dependent (controlled) sources have their output voltage or current determined by a voltage or current elsewhere in the circuit. There are four types:
1. VCVS: Voltage-Controlled Voltage Source (\(v = \mu v_x\))
2. CCVS: Current-Controlled Voltage Source (\(v = r i_x\))
3. VCCS: Voltage-Controlled Current Source (\(i = g v_x\))
4. CCCS: Current-Controlled Current Source (\(i = \beta i_x\))

The Supernode Technique

When an ideal voltage source (independent or dependent) is connected between two non-reference nodes:
1. Enclose the voltage source and its two bounding nodes into a generalized boundary called a supernode.
2. Apply KCL to the external terminals of this supernode.
3. Supplement with the internal constraint equation: \(v_{\text{pos}} - v_{\text{neg}} = V_s\).

The Supermesh Technique

When an ideal current source (independent or dependent) is shared between two adjacent meshes:
1. Temporarily remove the branch containing the current source to form a combined supermesh.
2. Write KVL around the perimeter of the supermesh.
3. Supplement with the internal branch constraint equation: \(i_{\text{mesh, aiding}} - i_{\text{mesh, opposing}} = I_s\).

1.4 Matrix Formulation in Network Analysis

For linear networks, mesh equations are formulated systematically as: \[\begin{bmatrix} R_{11} & R_{12} & \dots & R_{1n} \\ R_{21} & R_{22} & \dots & R_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ R_{n1} & R_{n2} & \dots & R_{nn} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ \vdots \\ I_n \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \\ \vdots \\ V_n \end{bmatrix}\] Where:
• \(R_{kk}\) = Self-resistance of mesh \(k\) (sum of all resistances in mesh \(k\)).
• \(R_{jk} = -(\text{Mutual resistance shared between mesh } j \text{ and } k)\).
• \(V_k\) = Algebraic sum of source voltages in mesh \(k\) aiding mesh current direction.

Impact of Dependent Sources on Matrix Symmetry

In purely passive networks, \([R]\) is always symmetrical (\(R_{jk} = R_{kj}\)). However, dependent sources introduce non-symmetrical off-diagonal terms once the controlling variable is substituted. Always move dependent terms to the left-hand matrix before solving!

1.5, 1.6 & 1.7 Procedure for Evaluating Initial Conditions & Derivatives

Universal Step-by-Step Procedure at Switching

  1. Analyze at \(t = 0^-\) (Pre-switching steady state):
    Under DC excitation for a long time:
    • All inductors behave as short circuits (\(v_L = 0\)). Find initial currents \(i_L(0^-)\).
    • All capacitors behave as open circuits (\(i_C = 0\)). Find initial voltages \(v_C(0^-)\).
  2. Apply Continuity Rules at \(t = 0^+\) (Post-switching instant): \[i_L(0^+) = i_L(0^-), \quad v_C(0^+) = v_C(0^-)\]
  3. Draw the Equivalent Circuit at \(t = 0^+\):
    • Replace each inductor with an ideal current source of value \(i_L(0^+)\) (if \(i_L(0^+) = 0\), replace with an open circuit).
    • Replace each capacitor with an ideal voltage source of value \(v_C(0^+)\) (if \(v_C(0^+) = 0\), replace with a short circuit).
    • Solve purely resistive network for all currents \(i(0^+)\) and voltages \(v(0^+)\).
  4. Evaluate First Derivatives \(\frac{di_L}{dt}(0^+)\) and \(\frac{dv_C}{dt}(0^+)\):
    From element v-i relations: \[\frac{di_L}{dt}(0^+) = \frac{v_L(0^+)}{L}, \quad \frac{dv_C}{dt}(0^+) = \frac{i_C(0^+)}{C}\] Write loop KVL or node KCL at \(t = 0^+\) to find \(v_L(0^+)\) and \(i_C(0^+)\).
  5. Evaluate Second Derivatives \(\frac{d^2 i}{dt^2}(0^+)\):
    Write KVL/KCL equations as general time-varying differential functions for \(t > 0\). Differentiate once with respect to \(t\), then evaluate at \(t = 0^+\) by substituting previously calculated initial values and first derivatives.

Worked Master Example (IOE Exam Problem)

Initial Conditions & Derivatives in Two-Mesh RLC Network PYQ 2081 Ashwin & 2083 Baishakh Q2
Problem Statement:
In the circuit, a 10V DC source is switched at \(t = 0\) into Mesh 1 with inductor \(L = 1\text{H}\) in series with switch and a shunt resistor of \(10\,\Omega\). Mesh 2 contains a series resistor of \(10\,\Omega\) and capacitor \(C = 1\text{F}\). Find \(i_1(0^+), i_2(0^+), \frac{di_1}{dt}(0^+), \frac{di_2}{dt}(0^+)\).
Step 1: State at \(t = 0^-\)
Prior to switching, circuit is completely unenergized: \[i_L(0^-) = 0\text{ A}, \quad v_C(0^-) = 0\text{ V}\]
Step 2: State at \(t = 0^+\)
By continuity: \[i_1(0^+) = i_L(0^+) = i_L(0^-) = 0\text{ A}\] \[v_C(0^+) = v_C(0^-) = 0\text{ V}\] KVL around Mesh 2 at \(t = 0^+\): \[10[i_2(0^+) - i_1(0^+)] + 10 i_2(0^+) + v_C(0^+) = 0\] Since \(i_1(0^+) = 0\) and \(v_C(0^+) = 0\): \[20 i_2(0^+) = 0 \implies i_2(0^+) = 0\text{ A}\]
Step 3: Calculating First Derivative \(\frac{di_1}{dt}(0^+)\)
KVL for Mesh 1 for \(t > 0\): \[10 = L \frac{di_1}{dt} + 10(i_1 - i_2)\] At \(t = 0^+\): \[10 = 1 \cdot \frac{di_1}{dt}(0^+) + 10(0 - 0) \implies \frac{di_1}{dt}(0^+) = 10\text{ A/s}\]
Step 4: Calculating First Derivative \(\frac{di_2}{dt}(0^+)\)
KVL for Mesh 2 for \(t > 0\): \[20 i_2 - 10 i_1 + v_C = 0\] Differentiate with respect to \(t\): \[20 \frac{di_2}{dt} - 10 \frac{di_1}{dt} + \frac{dv_C}{dt} = 0\] Since \(\frac{dv_C}{dt}(0^+) = \frac{i_C(0^+)}{C} = \frac{i_2(0^+)}{C} = \frac{0}{1} = 0\): \[20 \frac{di_2}{dt}(0^+) - 10(10) + 0 = 0 \implies 20 \frac{di_2}{dt}(0^+) = 100 \implies \frac{di_2}{dt}(0^+) = 5\text{ A/s}\]
Final Answer: \[i_1(0^+) = 0\text{ A}, \quad i_2(0^+) = 0\text{ A}, \quad \frac{di_1}{dt}(0^+) = 10\text{ A/s}, \quad \frac{di_2}{dt}(0^+) = 5\text{ A/s}\]

Common Mistakes to Avoid in Chapter 1

  • Assuming inductor current is always zero at \(t=0^+\): If a switch has been closed for a long time before moving at \(t=0\), the inductor will carry a non-zero steady-state current \(i_L(0^-) = I_0\) which persists at \(t=0^+\)!
  • Forgetting internal supernode constraint: In nodal analysis, forgetting to write the voltage difference relation (\(v_a - v_b = V_s\)) leaves the system with fewer equations than unknowns.
  • Differentiating constants incorrectly: Remember that a fixed DC voltage source has \(\frac{dV}{dt} = 0\).