Ch 1. Transients in Electric Circuit

7 hours · 7 marks

1.1 Characteristics of Various Network Elements

A lumped electric network is built from passive elements (R, L, C) and active sources (independent / dependent). Transient response arises because L and C store energy and cannot change that energy instantaneously.

Figure 1.1 — Passive circuit elements: R, L, C

Resistor (R)

Ohm’s law: \(v_R = Ri\), \(i_R = Gv\) with \(G = 1/R\).
Power (always dissipated): \(p = vi = Ri^2 = v^2/R\).
Memoryless: voltage and current can jump together; no energy storage.

Inductor (L)

\(v_L = L\dfrac{di}{dt}\), \(i_L = \dfrac{1}{L}\int_{-\infty}^{t} v\,d\tau = i(t_0) + \dfrac{1}{L}\int_{t_0}^{t} v\,d\tau\).
Energy stored: \(W_L = \tfrac12 Li^2\).
Continuity: inductor current is continuous for finite voltage: \(i_L(0^+) = i_L(0^-)\). An open-circuit inductor at \(t=0^+\) appears if \(i_L(0^-)=0\); a short for DC steady state (\(di/dt=0\)).

Capacitor (C)

\(i_C = C\dfrac{dv}{dt}\), \(v_C = v(t_0) + \dfrac{1}{C}\int_{t_0}^{t} i\,d\tau\).
Energy stored: \(W_C = \tfrac12 Cv^2\).
Continuity: capacitor voltage is continuous for finite current: \(v_C(0^+) = v_C(0^-)\). Uncharged capacitor looks like a short at \(t=0^+\); open circuit in DC steady state (\(dv/dt=0\)).

Independent sources

Independent voltage source: fixed \(v(t)\) regardless of load current (ideal \(R_s=0\)).
Independent current source: fixed \(i(t)\) regardless of terminal voltage (ideal \(R_p=\infty\)).

Dependent (controlled) sources

Type	Controlling variable	Controlled output	Symbol / gain
VCVS	Voltage \(v_x\)	Voltage \(\mu v_x\)	Voltage-controlled voltage source
VCCS	Voltage \(v_x\)	Current \(g_m v_x\)	Transconductance \(g_m\)
CCVS	Current \(i_x\)	Voltage \(r_m i_x\)	Transresistance \(r_m\)
CCCS	Current \(i_x\)	Current \(\beta i_x\)	Current gain \(\beta\)

Dependent sources model amplifiers/op-amps in circuit exams. Always express the controlling variable in terms of the same node voltages / mesh currents used in KCL/KVL.

Figure 1.2 — Four types of dependent sources

1.2 Nodal Analysis with Dependent and Independent Sources

Procedure:

Choose a reference (ground) node.
Assign unknown node voltages \(v_1,v_2,\ldots\) to remaining essential nodes.
Convert voltage sources in series with resistors to current sources if convenient (source transformation), or use a supernode for floating voltage sources.
Write KCL at each non-reference node: sum of currents leaving = 0 (or entering = leaving).
Express every branch current as \(G(v_a-v_b)\) plus independent/dependent source currents.
Substitute dependent-source laws (\(i = g_m v_x\), etc.) and solve the linear system.

General admittance form for \(n\) nodes:

\[ \mathbf{Y}\mathbf{V} = \mathbf{I}_{\text{s}} \] where \(Y_{ii}\) = sum of admittances connected to node \(i\), \(Y_{ij}=-G_{ij}\) for the branch between \(i\) and \(j\), and \(\mathbf{I}_s\) collects independent current injections (dependent parts move to the left side and modify \(\mathbf{Y}\)).

Worked Example 1 — Nodal analysis with VCCS

Nodes: reference, \(V_1\), \(V_2\). Elements: \(2\,\Omega\) from \(V_1\) to gnd, \(4\,\Omega\) between \(V_1\) and \(V_2\), \(4\,\Omega\) from \(V_2\) to gnd, independent current source \(3\,\text{A}\) into \(V_1\), and VCCS \(2V_1\) leaving node \(V_2\) toward gnd.

Figure 1.3 — Circuit for Worked Example 1 (Nodal analysis with VCCS)

KCL at \(V_1\):

\[ \frac{V_1}{2} + \frac{V_1-V_2}{4} = 3 \quad\Rightarrow\quad 2V_1 + (V_1-V_2) = 12 \quad\Rightarrow\quad 3V_1 - V_2 = 12 \tag{1} \]

KCL at \(V_2\):

\[ \frac{V_2-V_1}{4} + \frac{V_2}{4} + 2V_1 = 0 \quad\Rightarrow\quad V_2-V_1 + V_2 + 8V_1 = 0 \quad\Rightarrow\quad 7V_1 + 2V_2 = 0 \tag{2} \]

From (2): \(V_2 = -\tfrac72 V_1\). Substitute in (1):

\[ 3V_1 - \Bigl(-\frac{7}{2}V_1\Bigr)=12 \Rightarrow \Bigl(3+\frac{7}{2}\Bigr)V_1=12 \Rightarrow \frac{13}{2}V_1=12 \Rightarrow V_1=\frac{24}{13}\,\text{V},\quad V_2=-\frac{84}{13}\,\text{V}. \]

Final answer: \(V_1 = 24/13\,\text{V}\), \(V_2 = -84/13\,\text{V}\).

Worked Example 2 — Nodal analysis with CCVS (supernode idea)

A CCVS \(v = 4i_x\) sits between nodes \(V_a\) and \(V_b\) (positive at \(V_a\)), with controlling current \(i_x\) through a \(2\,\Omega\) sense resistor from \(V_b\) to ground. An independent \(10\,\text{V}\) source feeds \(V_a\) through \(1\,\Omega\).

Constraint from the CCVS:

\[ V_a - V_b = 4i_x,\qquad i_x = \frac{V_b}{2} \Rightarrow V_a - V_b = 2V_b \Rightarrow V_a = 3V_b. \]

KCL at the supernode \(\{V_a,V_b\}\) (current through the ideal CCVS cancels internally):

\[ \frac{V_a-10}{1} + \frac{V_b}{2} = 0 \Rightarrow V_a + \frac{V_b}{2} = 10. \] With \(V_a=3V_b\): \(3V_b + V_b/2 = 10 \Rightarrow (7/2)V_b=10 \Rightarrow V_b=20/7\,\text{V}\), \(V_a=60/7\,\text{V}\), \(i_x=10/7\,\text{A}\).

Final answer: \(V_a=60/7\,\text{V}\), \(V_b=20/7\,\text{V}\), \(i_x=10/7\,\text{A}\).

1.3 Mesh Analysis with Dependent and Independent Sources

Procedure:

Identify \(m\) independent meshes (planar circuit).
Assign clockwise mesh currents \(I_1,I_2,\ldots,I_m\).
Write KVL around each mesh: sum of voltage drops = 0.
Shared resistor \(R\) between meshes \(i\) and \(j\) contributes \(R(I_i-I_j)\) in mesh \(i\).
Current sources: if a current source is on the periphery, that mesh current is known; if shared by two meshes, form a supermesh and add the current-source constraint \(I_i-I_j = I_s\).
Express dependent sources using mesh currents, then solve \(\mathbf{Z}\mathbf{I}=\mathbf{V}\).

Worked Example 3 — Mesh + dependent CCVS, find \(V_x\) and power

Two meshes. Left mesh: \(12\,\text{V}\) source, \(2\,\Omega\), \(4\,\Omega\) shared. Right mesh: \(4\,\Omega\) shared, \(6\,\Omega\), and CCVS \(3I_1\) (polarity opposing \(I_2\)). \(V_x\) is voltage across the \(6\,\Omega\) resistor.

Mesh 1 KVL:

\[ -12 + 2I_1 + 4(I_1-I_2)=0 \Rightarrow 6I_1 - 4I_2 = 12 \Rightarrow 3I_1 - 2I_2 = 6 \tag{A} \]

Mesh 2 KVL:

\[ 4(I_2-I_1) + 6I_2 + 3I_1 = 0 \Rightarrow -4I_1 + 10I_2 + 3I_1 = 0 \Rightarrow -I_1 + 10I_2 = 0 \Rightarrow I_1 = 10I_2 \tag{B} \] Substitute (B) into (A): \(3(10I_2)-2I_2=6\Rightarrow 28I_2=6\Rightarrow I_2=3/14\,\text{A}\), \(I_1=15/7\,\text{A}\).

\[ V_x = 6I_2 = 6\cdot\frac{3}{14}=\frac{9}{7}\,\text{V}. \] Power absorbed by \(6\,\Omega\): \(P=I_2^2\cdot 6 = (9/196)\cdot 6 = 27/98\,\text{W}\).

Final answer: \(I_1=15/7\,\text{A}\), \(I_2=3/14\,\text{A}\), \(V_x=9/7\,\text{V}\), \(P_{6\Omega}=27/98\,\text{W}\).

1.4 Application of Matrix Method in Network Analysis

Both nodal and mesh equations are written compactly as \(\mathbf{A}\mathbf{x}=\mathbf{b}\) and solved by Cramer’s rule, Gaussian elimination, or \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\).

Mesh impedance matrix

\[ \begin{bmatrix} R_{11} & R_{12} & \cdots \\ R_{21} & R_{22} & \cdots \\ \vdots & \vdots & \ddots \end{bmatrix} \begin{bmatrix} I_1\\ I_2\\ \vdots \end{bmatrix} = \begin{bmatrix} V_1\\ V_2\\ \vdots \end{bmatrix} \]

\(R_{ii}\) = sum of all resistances in mesh \(i\).
\(R_{ij}=R_{ji}=-R_{\text{shared}}\) if both mesh currents enter the shared resistor from opposite sides (standard clockwise convention).
Dependent sources alter off-diagonal / diagonal entries after the controlling variable is substituted.

Worked Example 4 — \(2\times 2\) matrix for \(V_x\) and source power

From Example 3 before substituting the CCVS into numbers, the unmodified passive matrix would be

\[ \begin{bmatrix} 6 & -4 \\ -4 & 10 \end{bmatrix} \begin{bmatrix} I_1\\ I_2 \end{bmatrix} = \begin{bmatrix} 12\\ -3I_1 \end{bmatrix}. \] Move \(-3I_1\) to the left:

\[ \begin{bmatrix} 6 & -4 \\ -1 & 10 \end{bmatrix} \begin{bmatrix} I_1\\ I_2 \end{bmatrix} = \begin{bmatrix} 12\\ 0 \end{bmatrix}. \] Determinant \(\Delta=60-4=56\). By Cramer:

\[ I_1=\frac{\begin{vmatrix}12&-4\\0&10\end{vmatrix}}{56}=\frac{120}{56}=\frac{15}{7},\quad I_2=\frac{\begin{vmatrix}6&12\\-1&0\end{vmatrix}}{56}=\frac{12}{56}=\frac{3}{14}. \] Source power delivered by the \(12\,\text{V}\) source: \(P_s=12\cdot I_1=12\cdot 15/7=180/7\,\text{W}\).

Final answer: \(I_1=15/7\,\text{A}\), \(I_2=3/14\,\text{A}\), \(P_s=180/7\,\text{W}\).

1.5 Procedure of Evaluating Initial Conditions

Initial conditions are values of capacitor voltages and inductor currents (and their derivatives if needed) at the switching instant \(t=0^+\) (or \(t=t_0^+\)).

Steady state before switching (\(t=0^-\)): replace capacitors by open circuits, inductors by short circuits; solve the resulting resistive circuit for \(v_C(0^-)\) and \(i_L(0^-)\).
Continuity: \(v_C(0^+)=v_C(0^-)\), \(i_L(0^+)=i_L(0^-)\). (Exception: impulse sources / infinite \(dv\) or \(di\) — rare in IOE papers.)
Redraw circuit at \(t=0^+\): replace each capacitor by a DC voltage source of value \(v_C(0^+)\); replace each inductor by a DC current source of value \(i_L(0^+)\).
Solve this resistive circuit for any other requested quantities: resistor voltages/currents, \(v_L(0^+)\), \(i_C(0^+)\), source currents, etc.
If derivatives are required, use element laws: \(L\,di_L/dt|_{0^+}=v_L(0^+)\), \(C\,dv_C/dt|_{0^+}=i_C(0^+)\), and differentiate KCL/KVL if higher derivatives appear.

1.6 Initial Values of Derivatives

Exams often ask \(di/dt|_{0^+}\) or \(dv/dt|_{0^+}\) in addition to \(i(0^+)\), \(v(0^+)\).

For an inductor: \(\displaystyle\left.\frac{di_L}{dt}\right|_{0^+}=\frac{v_L(0^+)}{L}\).
For a capacitor: \(\displaystyle\left.\frac{dv_C}{dt}\right|_{0^+}=\frac{i_C(0^+)}{C}\).
Differentiate the differential equation of the network and evaluate at \(t=0^+\) using already-known \(0^+\) values when second derivatives are needed (e.g. for series RLC: \(L\,d^2i/dt^2 + R\,di/dt + i/C = dv_s/dt\)).

Worked Example 5 — \(i\), \(di/dt\), \(v\), \(dv/dt\) at \(t=0^+\)

Series \(R=4\,\Omega\), \(L=1\,\text{H}\), \(C=1/4\,\text{F}\) with DC source \(12\,\text{V}\) switched in at \(t=0\). Before \(t=0\) the circuit is relaxed (\(i_L(0^-)=0\), \(v_C(0^-)=0\)).

At \(t=0^+\): \(i(0^+)=i_L(0^+)=0\), \(v_C(0^+)=0\). KVL at \(0^+\):

\[ 12 = Ri(0^+) + v_L(0^+) + v_C(0^+) = v_L(0^+) \Rightarrow v_L(0^+)=12\,\text{V} \Rightarrow \left.\frac{di}{dt}\right|_{0^+}=\frac{12}{L}=12\,\text{A/s}. \] Capacitor current equals series current, so \(i_C(0^+)=0\) and

\[ \left.\frac{dv_C}{dt}\right|_{0^+}=\frac{i_C(0^+)}{C}=0. \] (Immediately after switching the uncharged capacitor still has zero voltage and zero current because inductor current is still zero; voltage starts rising as current builds.)

Final answer: \(i(0^+)=0\), \(di/dt|_{0^+}=12\,\text{A/s}\), \(v_C(0^+)=0\), \(dv_C/dt|_{0^+}=0\).

1.7 Initial Condition in R-L-C Network

In a general RLC network after switching:

Keep all independent sources at their \(t=0^+\) values (step sources jump; sinusoids take \(\sin/\cos\) at \(0^+\)).
Replace \(C\) by voltage source \(v_C(0^+)\), \(L\) by current source \(i_L(0^+)\).
Solve for branch currents/voltages; then get derivatives from \(v_L=L\,di/dt\), \(i_C=C\,dv/dt\).

Worked Example 6 — Switched RL–C with prior steady state

For \(t<0\): a \(24\,\text{V}\) DC source, \(6\,\Omega\) resistor, and \(2\,\text{H}\) inductor in series have been connected long enough; capacitor \(C=1/3\,\text{F}\) across the inductor+resistor combination is open in DC steady state. Switch rearranges at \(t=0\) so that the source is removed and a series RLC discharge loop of \(R=6\,\Omega\), \(L=2\,\text{H}\), \(C=1/3\,\text{F}\) remains with initial energy on \(C\) and \(L\).

Before switching: inductor short \(\Rightarrow\) \(i_L(0^-)=24/6=4\,\text{A}\). Capacitor sees full \(24\,\text{V}\) (across the shorted L and the path) depending on topology — assume parallel C across the series RL so \(v_C(0^-)=24\,\text{V}\).

Thus \(i_L(0^+)=4\,\text{A}\), \(v_C(0^+)=24\,\text{V}\). At \(t=0^+\) in the discharge loop KVL:

\[ Ri(0^+) + v_L(0^+) + v_C(0^+) = 0 \Rightarrow 6\cdot 4 + v_L(0^+) + 24 = 0 \Rightarrow v_L(0^+)=-48\,\text{V} \Rightarrow \left.\frac{di}{dt}\right|_{0^+}=\frac{-48}{2}=-24\,\text{A/s}. \] \[ i_C(0^+)=-i_L(0^+)=-4\,\text{A (sign by discharge direction)},\quad \left.\frac{dv_C}{dt}\right|_{0^+}=\frac{-4}{1/3}=-12\,\text{V/s}. \]

Final answer: \(i(0^+)=4\,\text{A}\), \(di/dt|_{0^+}=-24\,\text{A/s}\), \(v_C(0^+)=24\,\text{V}\), \(dv_C/dt|_{0^+}=-12\,\text{V/s}\).

Most Probable Questions

Q1. State the continuity properties of inductor current and capacitor voltage. Why can resistor voltage jump at switching? 3 marks

\(i_L(0^+)=i_L(0^-)\) for finite \(v_L\) because \(\Delta i = (1/L)\int v\,dt\) stays finite over an infinitesimal interval. \(v_C(0^+)=v_C(0^-)\) similarly. A resistor has no storage; its \(v=Ri\) follows whatever current the L/C constraints allow, so it may jump.

Q2. Classify the four types of dependent sources with symbols for their gains. 4 marks

VCVS (\(\mu\)), VCCS (\(g_m\)), CCVS (\(r_m\)), CCCS (\(\beta\)). Controlling variable is either a branch voltage or current; controlled quantity is either voltage or current (see §1.1 table).

Q3. Using nodal analysis, find node voltages in a circuit containing a VCCS. 7 marks

Write KCL at each node; express the VCCS current as \(g_m v_x\) with \(v_x\) in terms of node voltages; solve the linear system (see Worked Example 1).

Q4. Write mesh equations in matrix form for a two-mesh circuit with a CCVS and find \(V_x\) and power. 7 marks

Form \(\mathbf{Z}\mathbf{I}=\mathbf{V}\), move dependent source terms into \(\mathbf{Z}\), solve by determinants, then \(V_x=R I_k\) and \(P=I^2R\) or \(VI\) (see Examples 3–4).

Q5. Outline the step-by-step procedure to evaluate initial conditions in an RLC network. 5 marks

Steady state at \(0^-\) (C open, L short) → continuity of \(v_C\), \(i_L\) → redraw at \(0^+\) with C as voltage source and L as current source → solve resistive circuit → derivatives via \(v_L/L\) and \(i_C/C\) (§1.5–1.6).

Q6. For a series RLC switched to a DC source at \(t=0\) with zero initial energy, find \(i(0^+)\), \(di/dt|_{0^+}\), \(v_C(0^+)\), \(dv_C/dt|_{0^+}\). 7 marks

\(i(0^+)=0\), \(v_C(0^+)=0\), \(v_L(0^+)=V_s\), \(di/dt|_{0^+}=V_s/L\), \(dv_C/dt|_{0^+}=0\) (Example 5). Show KVL clearly for full marks.

Q7. Explain how dependent sources modify the bus admittance / mesh impedance matrix. 4 marks

Independent sources appear only in \(\mathbf{b}\). After substituting \(i=\beta i_x\) or \(v=\mu v_x\), terms proportional to unknowns move to the left, changing diagonal and off-diagonal entries of \(\mathbf{Y}\) or \(\mathbf{Z}\) (asymmetry is possible).

Q8. In a network that was in DC steady state, a switch opens at \(t=0\). How do you find \(v_L(0^+)\) and \(di_L/dt|_{0^+}\)? 5 marks

Find \(i_L(0^-)\) with L shorted; set \(i_L(0^+)=i_L(0^-)\); replace L by that current source in the \(t=0^+\) circuit; compute voltage across the inductor terminals \(v_L(0^+)\); then \(di_L/dt|_{0^+}=v_L(0^+)/L\).

Next: Chapter 2 — Transient Analysis (Classical) →