Official IOE Past Question Papers

PYQ Question Bank & Verified Step-by-Step Solutions

Rigorous analysis of all 4 official examination papers of the new ENEE 154 syllabus (2081 Ashwin through 2083 Baishakh). Every numerical calculation is completely verified using the standard 8-step engineering method.

Systematic Pattern & Trend Analysis Across 4 Exam Papers

1. Exact Question Duplication 100% Match

Q2 of 2081 Ashwin and Q2 of 2083 Baishakh are word-for-word, value-for-value identical: Two-mesh circuit with \(10\text{V}\) source, \(1\text{H}\) inductor, two \(10\,\Omega\) resistors, and \(1\text{F}\) capacitor finding \(i_1(0^+), i_2(0^+), di_1/dt(0^+), di_2/dt(0^+)\).

2. Core Flux Independence 3 out of 4 Papers

The conceptual proof showing that "main magnetic flux in transformer core is independent of load current" appeared in 2081 Ashwin (Q7), 2082 Bhadra (Q10b), and 2083 Baishakh (Q7).

3. Asymptotic Bode Plots 4 out of 4 Papers

Always evaluated for 7 or 8 marks:
• 2083 Baishakh: \(H(s) = \frac{10(s^2+8s+15)}{s^2(s+3)(0.05s+1)} = \frac{10(s+5)}{s^2(0.05s+1)}\)
• 2082 Bhadra: \(T(s) = \frac{100(s+5)}{s^2(s^2+21s+20)} = \frac{100(s+5)}{s^2(s+1)(s+20)}\)
• 2082 Baishakh: \(H(s) = \frac{100s(s+5)}{(s+1)(s^2+10s+100)}\)
• 2081 Ashwin: \(G(s) = \frac{500(s+2)}{s(s^2+15s+121)}\)

4. Synchronous Speed Paradox Repeated Concept

"Why does the speed of a three-phase induction motor never reach synchronous speed?" appeared as core conceptual justification in 2082 Bhadra (Q10a) and 2082 Baishakh (Q9).

TRIBHUVAN UNIVERSITY • IOE EXAMINATION

2083 Baishakh — Back Exam (New Course)

Program: BEI, BCT | Year/Part: I / II | Full Marks: 60 | Pass Marks: 24 | Time: 3 hrs

Question 1 [6 Marks]: Nodal Analysis with Dependent Source

Question Statement: Find the current in the 10V source using nodal analysis for the circuit with node voltages \(v_1, v_2, v_3\). A \(6\,\Omega\) resistor connects \(v_1\) and \(v_3\). A \(2\,\Omega\) resistor connects \(v_1\) to ground with downward current \(i = v_1 / 2\). A 10V independent voltage source connects between \(v_1\) and \(v_2\) (with '+' at \(v_2\), '-' at \(v_1\)). A \(4\,\Omega\) resistor connects \(v_2\) to ground. A dependent voltage source of \(5i\) connects between \(v_2\) and \(v_3\) (with '+' at \(v_2\), '-' at \(v_3\)), and a \(3\,\Omega\) resistor connects \(v_3\) to ground.

Step-by-Step Nodal Solution Chapter 1

Step 1: Constraint Equations & Supernodes

Control variable: \(i = \frac{v_1}{2}\).
Between \(v_1\) and \(v_2\): \(v_2 - v_1 = 10 \implies v_2 = v_1 + 10\).
Between \(v_2\) and \(v_3\): \(v_2 - v_3 = 5i = 5\left(\frac{v_1}{2}\right) = 2.5 v_1 \implies v_3 = v_2 - 2.5 v_1 = (v_1 + 10) - 2.5 v_1 = 10 - 1.5 v_1\).
Notice that nodes 1, 2, and 3 are all interconnected by ideal voltage sources (one independent, one dependent), forming a single combined supernode enclosed around nodes 1, 2, and 3!

Step 2: KCL at Supernode (Nodes 1-2-3)

Sum of all currents leaving the supernode to ground: \[\frac{v_1}{2} + \frac{v_2}{4} + \frac{v_3}{3} = 0\] (The \(6\,\Omega\) resistor connects between \(v_1\) and \(v_3\), which is strictly internal to the supernode, so its current cancels out in the supernode KCL!).

Step 3: Substitution & Algebraic Solution

Substitute \(v_2 = v_1 + 10\) and \(v_3 = 10 - 1.5 v_1\): \[\frac{v_1}{2} + \frac{v_1 + 10}{4} + \frac{10 - 1.5 v_1}{3} = 0\] Multiply entire equation by 12: \[6 v_1 + 3(v_1 + 10) + 4(10 - 1.5 v_1) = 0\] \[6 v_1 + 3 v_1 + 30 + 40 - 6 v_1 = 0 \implies 3 v_1 + 70 = 0 \implies v_1 = -\frac{70}{3} \approx -23.33\text{ V}\] Now compute \(v_2\) and \(v_3\): \[v_2 = v_1 + 10 = -\frac{70}{3} + \frac{30}{3} = -\frac{40}{3} \approx -13.33\text{ V}\] \[v_3 = 10 - 1.5\left(-\frac{70}{3}\right) = 10 + 35 = 45\text{ V}\] Control current: \(i = \frac{v_1}{2} = -\frac{35}{3} \approx -11.67\text{ A}\).

Step 4: Current in the 10V Source

Apply KCL strictly at Node 1 to find current \(I_{10V}\) flowing from node 1 to node 2: \[I_{10V} + \frac{v_1}{2} + \frac{v_1 - v_3}{6} = 0\] \[I_{10V} = -\left[\frac{-70/3}{2} + \frac{-70/3 - 45}{6}\right] = -\left[-\frac{35}{3} + \frac{-205/3}{6}\right] = -\left[-\frac{70}{6} - \frac{205}{18}\right] = \frac{210 + 205}{18} = \frac{415}{18} \approx 23.06\text{ A}\]

Final Answer: Node voltages are \(v_1 = -23.33\text{ V}\), \(v_2 = -13.33\text{ V}\), \(v_3 = 45.0\text{ V}\). The current flowing through the 10V source (from \(v_1\) to \(v_2\)) is \(23.06\text{ A}\).

Question 2 [5 Marks]: Initial Condition Values at \(t = 0^+\) (Repeated in 2081 Ashwin)

Question Statement: Obtain the values of \(i_1(0^+), i_2(0^+), \frac{di_1}{dt}(0^+), \frac{di_2}{dt}(0^+)\), if the switch is closed at \(t = 0\) in the circuit shown: A 10V DC source connected via a switch to mesh 1 containing an inductor \(L = 1\text{H}\) in the top branch, a shunt resistor of \(10\,\Omega\), and mesh 2 containing a series resistor \(10\,\Omega\) and capacitor \(C = 1\text{F}\).

Step-by-Step Initial Conditions Evaluation Chapter 1

Step 1: State at \(t = 0^-\)

Before \(t=0\), the switch was open and the circuit was unenergized: \[i_L(0^-) = 0 \text{ A}, \quad v_C(0^-) = 0 \text{ V}\]

Step 2: Switching Invariance at \(t = 0^+\)

Current through an inductor and voltage across a capacitor cannot change instantaneously: \[i_L(0^+) = i_L(0^-) = 0 \implies i_1(0^+) = 0\text{ A}\] \[v_C(0^+) = v_C(0^-) = 0\text{ V}\] At \(t=0^+\), the inductor is in series with mesh 1, so \(i_1(0^+) = 0\).
For mesh 2, KVL gives: \(10(i_2(0^+) - i_1(0^+)) + 10 i_2(0^+) + v_C(0^+) = 0\).
Since \(i_1(0^+) = 0\) and \(v_C(0^+) = 0\): \[20 i_2(0^+) = 0 \implies i_2(0^+) = 0\text{ A}\]

Step 3: Evaluating \(\frac{di_1}{dt}(0^+)\)

Write KVL for Mesh 1 for \(t > 0\): \[10 = L \frac{di_1}{dt} + 10(i_1 - i_2)\] At \(t = 0^+\), substitute \(L = 1\text{H}\), \(i_1(0^+) = 0\), \(i_2(0^+) = 0\): \[10 = 1 \cdot \frac{di_1}{dt}(0^+) + 10(0 - 0) \implies \frac{di_1}{dt}(0^+) = 10\text{ A/s}\]

Step 4: Evaluating \(\frac{di_2}{dt}(0^+)\)

Write KVL for Mesh 2 for \(t > 0\): \[10(i_2 - i_1) + 10 i_2 + v_C = 0 \implies 20 i_2 - 10 i_1 + v_C = 0\] Differentiate with respect to \(t\): \[20 \frac{di_2}{dt} - 10 \frac{di_1}{dt} + \frac{dv_C}{dt} = 0\] Since \(\frac{dv_C}{dt} = \frac{i_C}{C} = \frac{i_2}{C}\), at \(t = 0^+\): \[\frac{dv_C}{dt}(0^+) = \frac{i_2(0^+)}{1} = \frac{0}{1} = 0\text{ V/s}\] Substitute into differentiated KVL: \[20 \frac{di_2}{dt}(0^+) - 10(10) + 0 = 0 \implies 20 \frac{di_2}{dt}(0^+) = 100 \implies \frac{di_2}{dt}(0^+) = 5\text{ A/s}\]

Final Results: \[i_1(0^+) = 0\text{ A}, \quad i_2(0^+) = 0\text{ A}, \quad \frac{di_1}{dt}(0^+) = 10\text{ A/s}, \quad \frac{di_2}{dt}(0^+) = 5\text{ A/s}\]

Question 3 [6 Marks]: Classical Transient Derivation (Series RC with Exponential Source)

Question Statement: Derive the expression of current in series RC circuit excited by exponential voltage source by using classical method.

See complete step-by-step mathematical proof in the Derivation Sheet Section 5.

Final Expression:

\[i(t) = \frac{V_0}{R(1 - a RC)} \left[ e^{-t/RC} - a RC \, e^{-at} \right]\]

Question 4 [6 Marks]: Laplace Transform Method for RL/RC Circuit

Question Statement: In the given circuit, switch is opened at \(t = 0\). Find the current and voltage across inductor using Laplace transform method. Circuit: 4V DC source connected in parallel with a \(1\text{F}\) capacitor, and a branch with inductor \(L = 0.5\text{H}\) in series with \(R = 1\,\Omega\).

Laplace Transform Solution Chapter 3

Step 1: Steady-State Initial Values at \(t = 0^-\)

Before \(t = 0\), the switch was closed for a long time. In DC steady state:
• Capacitor acts as open circuit: \(v_C(0^-) = 4\text{ V}\).
• Inductor acts as short circuit: \(i_L(0^-) = \frac{4\text{ V}}{1\,\Omega} = 4\text{ A}\).

Step 2: s-Domain Transformed Circuit for \(t > 0\)

At \(t = 0\), switch opens, disconnecting the 4V source. The circuit becomes a single parallel loop of capacitor discharging into the series \(R-L\) branch.
s-Domain representation:
• Capacitor: impedance \(\frac{1}{sC} = \frac{1}{s}\) with initial series voltage source \(\frac{v_C(0^-)}{s} = \frac{4}{s}\).
• Inductor: impedance \(sL = 0.5s\) with initial series voltage generator \(L i_L(0^-) = 0.5 \times 4 = 2\text{ V}\) opposing current drop.
• Resistor: \(R = 1\,\Omega\).

Step 3: Solving for \(I(s)\)

Applying KVL in the loop for clockwise current \(I(s)\): \[\frac{4}{s} - \frac{1}{s} I(s) - 1 \cdot I(s) - 0.5 s I(s) + 2 = 0\] \[I(s) \left( 0.5 s + 1 + \frac{1}{s} \right) = \frac{4}{s} + 2 = \frac{2s + 4}{s}\] \[I(s) \left( \frac{0.5 s^2 + s + 1}{s} \right) = \frac{2s + 4}{s} \implies I(s) = \frac{2s + 4}{0.5 s^2 + s + 1} = \frac{4(s + 2)}{s^2 + 2s + 2}\]

Step 4: Inverse Laplace Transform

Complete the square in denominator: \(s^2 + 2s + 2 = (s + 1)^2 + 1\). \[I(s) = \frac{4(s + 1) + 4}{(s + 1)^2 + 1} = 4 \frac{s + 1}{(s + 1)^2 + 1^2} + 4 \frac{1}{(s + 1)^2 + 1^2}\] Taking the inverse Laplace transform: \[i_L(t) = 4 e^{-t} \cos(t) + 4 e^{-t} \sin(t) = 4 e^{-t} (\cos t + \sin t)\text{ A} \quad (t \ge 0)\] Voltage across inductor: \[v_L(t) = L \frac{di_L}{dt} = 0.5 \frac{d}{dt}\left[ 4 e^{-t}(\cos t + \sin t) \right]\] \[\frac{di_L}{dt} = 4 \left[ -e^{-t}(\cos t + \sin t) + e^{-t}(-\sin t + \cos t) \right] = 4 e^{-t} (-2\sin t) = -8 e^{-t} \sin t\] \[v_L(t) = 0.5 (-8 e^{-t} \sin t) = -4 e^{-t} \sin(t)\text{ V}\]

Final Results: \[i_L(t) = 4 e^{-t} (\cos t + \sin t)\text{ A}, \quad v_L(t) = -4 e^{-t} \sin(t)\text{ V}\]

Question 5 [7 Marks]: Network Function & Asymptotic Bode Plot

Question Statement: What is a network function? Plot the asymptotic Bode graph of the transfer function: \[H(s) = \frac{10(s^2+8s+15)}{s^2(s+3)(0.05s+1)}\]

Bode Plot Construction Chapter 4

Step 1: Factorization & Cancellation

Factor the numerator quadratic: \(s^2 + 8s + 15 = (s + 3)(s + 5)\). \[H(s) = \frac{10(s+3)(s+5)}{s^2(s+3)(0.05s+1)} = \frac{10(s+5)}{s^2(0.05s+1)}\]

Step 2: Standard Time-Constant (Bode) Form

\[H(s) = \frac{10 \times 5 \left(1 + \frac{s}{5}\right)}{s^2 (1 + 0.05 s)} = \frac{50 \left(1 + \frac{s}{5}\right)}{s^2 \left(1 + \frac{s}{20}\right)}\] Bode gain \(K_B = 50\).
• Corner frequency for simple zero: \(\omega_{z1} = 5\text{ rad/s}\).
• Corner frequency for simple pole: \(\omega_{p1} = \frac{1}{0.05} = 20\text{ rad/s}\).
• Double pole at origin: \(1/s^2\).

Step 3: Magnitude Slope Table

Frequency Range (rad/s)	Contributing Factor	Slope Change	Net Slope
\(\omega < 5\)	Double pole at origin \(1/s^2\)	Initial	\(-40\text{ dB/decade}\)
\(5 \le \omega < 20\)	Zero at \(\omega = 5\)	\(+20\text{ dB/decade}\)	\(-20\text{ dB/decade}\)
\(\omega \ge 20\)	Pole at \(\omega = 20\)	\(-20\text{ dB/decade}\)	\(-40\text{ dB/decade}\)

Reference gain at \(\omega = 1\text{ rad/s}\): \[|H(j1)|_{\text{dB}} \approx 20\log_{10}(K_B) - 40\log_{10}(1) = 20\log_{10}(50) = 20(1.699) = 33.98\text{ dB}\] Gain at first corner frequency \(\omega = 5\): \[|H(j5)|_{\text{dB}} = 33.98 - 40\log_{10}(5) = 33.98 - 40(0.699) = 33.98 - 27.96 = 6.02\text{ dB}\]

Step 4: Phase Angle Equation

\[\phi(\omega) = -180^\circ + \tan^{-1}\left(\frac{\omega}{5}\right) - \tan^{-1}(0.05\omega)\]

Summary: Low-frequency slope is \(-40\text{ dB/dec}\); breaks at \(\omega = 5\text{ rad/s}\) to \(-20\text{ dB/dec}\); breaks at \(\omega = 20\text{ rad/s}\) back to \(-40\text{ dB/dec}\).

Question 6 [6 Marks]: Z-Parameters of Two-Port Network

Question Statement: Find Z parameter for given two port network and state whether network is symmetrical or reciprocal. Network structure: \(8\,\Omega\) resistor in series at Port 1, shunt resistor \(20\,\Omega\), series resistor \(10\,\Omega\), shunt resistor \(20\,\Omega\) across Port 2.

Z-Parameter Evaluation Chapter 5

Step 1: Open-Circuit at Port 2 (\(I_2 = 0\))

When port 2 is open:
Resistors \(10\,\Omega\) and \(20\,\Omega\) are in series (\(= 30\,\Omega\)), connected in parallel with the first \(20\,\Omega\) resistor: \[R_{p1} = 20 \parallel 30 = \frac{20 \times 30}{50} = 12\,\Omega\] Input impedance at Port 1: \[Z_{11} = \left.\frac{V_1}{I_1}\right|_{I_2 = 0} = 8 + R_{p1} = 8 + 12 = 20\,\Omega\] Voltage across the second \(20\,\Omega\) resistor: By voltage division across the parallel branch: \[V_2 = V_{20\Omega,\text{shunt}} \times \frac{20}{10 + 20} = (I_1 \times 12) \times \frac{2}{3} = 8 I_1 \implies Z_{21} = \left.\frac{V_2}{I_1}\right|_{I_2 = 0} = 8\,\Omega\]

Step 2: Open-Circuit at Port 1 (\(I_1 = 0\))

When port 1 is open, no current flows through the \(8\,\Omega\) resistor. Current \(I_2\) enters Port 2. The parallel branch across Port 2 is the shunt \(20\,\Omega\) in parallel with \((10 + 20 = 30\,\Omega)\): \[Z_{22} = \left.\frac{V_2}{I_2}\right|_{I_1 = 0} = 20 \parallel (10 + 20) = 12\,\Omega\] Voltage at Port 1 is the voltage across the first \(20\,\Omega\) resistor: \[V_1 = V_2 \times \frac{20}{10 + 20} = (12 I_2) \times \frac{2}{3} = 8 I_2 \implies Z_{12} = \left.\frac{V_1}{I_2}\right|_{I_1 = 0} = 8\,\Omega\]

Step 3: Checking Reciprocity & Symmetry

• Reciprocity: \(Z_{12} = Z_{21} = 8\,\Omega\). Thus, the network is RECIPROCAL.
• Symmetry: \(Z_{11} = 20\,\Omega \ne Z_{22} = 12\,\Omega\). Thus, the network is NOT SYMMETRICAL.

\[[Z] = \begin{bmatrix} 20 & 8 \\ 8 & 12 \end{bmatrix} \,\Omega\] Network is Reciprocal (\(Z_{12} = Z_{21}\)) but Asymmetrical (\(Z_{11} \ne Z_{22}\)).

Question 7 [5 Marks]: Features of Ideal Transformer & Core Flux Constancy

Features of Ideal Transformer:

Zero winding resistance (\(R_1 = R_2 = 0\), zero copper losses).
Infinite core permeability (\(\mu_r \to \infty\), zero magnetizing current required).
Zero core losses (no hysteresis or eddy current losses, \(P_i = 0\)).
Zero leakage flux (100% magnetic coupling, \(k = 1\), all flux is confined to core).

For the proof that core flux is independent of load current, see Derivations Sheet Section 2.

Question 8 [5 Marks]: DC Shunt Generator Generated EMF & Efficiency

DC Shunt Generator Numerical Chapter 8

Given:

Terminal voltage \(V_t = 200\text{ V}\)
Load current \(I_L = 96\text{ A}\)
Armature resistance \(R_a = 0.1\,\Omega\)
Shunt field resistance \(R_{sh} = 50\,\Omega\)
Iron and friction losses = \(2500\text{ W}\)

Required:

(a) Generated EMF \(E_g\)
(b) Commercial efficiency \(\eta\)

Principle & Formula:

\[I_{sh} = \frac{V_t}{R_{sh}}, \quad I_a = I_L + I_{sh}\] \[E_g = V_t + I_a R_a\] \[\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{V_t I_L}{V_t I_L + \text{Total Losses}} \times 100\%\]

Calculation:

1. Field current: \(I_{sh} = \frac{200}{50} = 4\text{ A}\).
2. Armature current: \(I_a = 96 + 4 = 100\text{ A}\).
3. Generated EMF: \[E_g = 200 + (100 \times 0.1) = 200 + 10 = 210\text{ V}\] 4. Power Output: \[P_{\text{out}} = V_t I_L = 200 \times 96 = 19,200\text{ W} = 19.2\text{ kW}\] 5. Copper Losses:
• Armature Cu loss: \(I_a^2 R_a = 100^2 \times 0.1 = 1000\text{ W}\).
• Shunt Cu loss: \(V_t I_{sh} = 200 \times 4 = 800\text{ W}\).
• Total Cu loss: \(1000 + 800 = 1800\text{ W}\).
6. Total Losses: \[\text{Total Losses} = \text{Cu Losses} + \text{Iron \& Friction} = 1800 + 2500 = 4300\text{ W}\] 7. Commercial Efficiency: \[\eta = \frac{19,200}{19,200 + 4,300} \times 100\% = \frac{19,200}{23,500} \times 100\% \approx 81.70\%\]

Final Results: (a) Generated EMF \(E_g = 210\text{ V}\), (b) Commercial Efficiency \(\eta = 81.70\%\).

Question 9 [5 Marks]: 3-Phase Induction Motor Slip & Rotor Frequency

Induction Motor Speed & Rotor Frequency Chapter 9

Given:

Poles \(P = 4\)
Supply frequency \(f = 50\text{ Hz}\)
Part (a) slip \(s = 5\% = 0.05\)
Part (b) rotor speed \(N = 600\text{ rpm}\)

Calculations:

Synchronous speed: \[N_s = \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500\text{ rpm}\] (a) Speed when slip is 5%: \[N = N_s (1 - s) = 1500 (1 - 0.05) = 1500 \times 0.95 = 1425\text{ rpm}\] (b) Rotor EMF frequency when motor runs at 600 rpm: \[s = \frac{N_s - N}{N_s} = \frac{1500 - 600}{1500} = \frac{900}{1500} = 0.6\] \[f_r = s f = 0.6 \times 50 = 30\text{ Hz}\]

Final Results: (a) Motor speed = \(1425\text{ rpm}\), (b) Rotor frequency at 600 rpm = \(30\text{ Hz}\).

Question 10 [9 Marks]: Short Notes (Series/Parallel Magnetic, Universal Motor, Starter, VR)

a) Series and Parallel Magnetic Circuits: In series magnetic circuits, the same flux \(\Phi\) passes through all sections; total \(\text{MMF} = \Phi(\mathcal{R}_1 + \mathcal{R}_2 + \mathcal{R}_g)\). In parallel circuits, total flux divides across branches: \(\Phi = \Phi_1 + \Phi_2\), and MMF across parallel paths is identical.

b) Universal Motor: A series-wound commutator motor designed to operate efficiently on either DC or single-phase AC. Operates at very high speeds (\(3000 - 15000\text{ rpm}\)), ideal for vacuum cleaners, blenders, and portable drills.

c) DC Motor Starter (3-Point Starter): Inserts starting resistors in series with armature to limit huge inrush currents (\(I_a = V/R_a\)). Incorporates a No-Volt Coil (NVC) to release the handle on power outage and an Overload Release (OLR) coil to trip on excessive load.

d) Voltage Regulation of Transformer: Defined as \(\%VR = \frac{V_{2,\text{no-load}} - V_{2,\text{full-load}}}{V_{2,\text{no-load}}} \times 100\). Measures the voltage drop due to internal resistance and leakage reactance.

TRIBHUVAN UNIVERSITY • IOE EXAMINATION

2082 Bhadra — Regular Exam (New Course)

Program: BEI, BCT | Year/Part: I / II | Full Marks: 60 | Pass Marks: 24 | Time: 3 hrs

Question 1 [6 Marks]: Mesh Analysis with Dependent Voltage Source

Question Statement: Find the current \(I_x\) and power delivered by the dependent voltage source in the circuit shown below using Mesh Analysis. Circuit: 100V independent DC source, dependent source \(10 I_x\), resistors \(5\,\Omega, 10\,\Omega, 15\,\Omega, 50\,\Omega, 25\,\Omega\).

Set up standard mesh KVL matrix \([R][I] = [V]\), substitute control variable \(I_x\) in terms of mesh currents, and solve by matrix inversion / Cramer's rule.

Question 2 [1+4 Marks]: Inductor Switching & Derivatives at \(t = 0^+\)

Question Statement: Explain the behavior of an inductor when subjected to switching. In the network, switch is opened at \(t = 0\). Calculate \(v, \frac{dv}{dt}, \frac{d^2v}{dt^2}\) at \(t = 0^+\) if \(I = 2\text{A}, R = 200\,\Omega, L = 1\text{H}\). Given that \(v\) is the voltage across the inductor.

Higher Derivatives Calculation Chapter 1

Calculations:

Before switching: Switch closed, inductor in steady state carries \(i_L(0^-) = 2\text{A}\).
At \(t = 0^+\): Switch opens. Current source \(I = 2\text{A}\) must now split between \(R\) and \(L\).
Since \(i_L(0^+) = 2\text{A}\), all current from the source enters the inductor, meaning resistor current \(i_R(0^+) = I - i_L(0^+) = 2 - 2 = 0\text{ A}\).
Therefore, inductor voltage \(v(0^+) = v_R(0^+) = i_R(0^+) \times R = 0 \times 200 = 0\text{ V}\).
For \(t > 0\): \(i_R + i_L = I \implies \frac{v}{R} + i_L = I\).
Differentiate with respect to \(t\): \[\frac{1}{R}\frac{dv}{dt} + \frac{di_L}{dt} = 0 \implies \frac{1}{R}\frac{dv}{dt} + \frac{v}{L} = 0 \implies \frac{dv}{dt} = -\frac{R}{L} v\] At \(t = 0^+\), since \(v(0^+) = 0\): \[\frac{dv}{dt}(0^+) = -\frac{200}{1}(0) = 0\text{ V/s}\] Differentiating again: \[\frac{d^2v}{dt^2} = -\frac{R}{L}\frac{dv}{dt} \implies \frac{d^2v}{dt^2}(0^+) = 0\text{ V/s}^2\]

Final Values: \(v(0^+) = 0\text{ V}\), \(\frac{dv}{dt}(0^+) = 0\text{ V/s}\), \(\frac{d^2v}{dt^2}(0^+) = 0\text{ V/s}^2\).

Question 6 [1+7 Marks]: Poles & Zeros and Asymptotic Bode Plot

Transfer Function:

\[T(s) = \frac{100(s+5)}{s^2(s^2+21s+20)} = \frac{100(s+5)}{s^2(s+1)(s+20)}\]

Bode Standard Form:

\[T(s) = \frac{100 \times 5 \left(1 + \frac{s}{5}\right)}{s^2 \left(1 + \frac{s}{1}\right) \times 20 \left(1 + \frac{s}{20}\right)} = \frac{25 \left(1 + \frac{s}{5}\right)}{s^2 (1 + s)\left(1 + \frac{s}{20}\right)}\]

Bode gain \(K_B = 25 \implies 20\log_{10}(25) = 27.96\text{ dB}\).
Corner frequencies: \(\omega_{p1} = 1\text{ rad/s}\) (pole), \(\omega_{z1} = 5\text{ rad/s}\) (zero), \(\omega_{p2} = 20\text{ rad/s}\) (pole).
Initial slope for \(\omega < 1\): \(-40\text{ dB/dec}\) (due to \(1/s^2\)).
At \(\omega = 1\): slope becomes \(-40 - 20 = -60\text{ dB/dec}\).
At \(\omega = 5\): slope becomes \(-60 + 20 = -40\text{ dB/dec}\).
At \(\omega = 20\): slope becomes \(-40 - 20 = -60\text{ dB/dec}\).

Question 8 [5 Marks]: Magnetic Circuit Air Gap MMF Numerical

Composite Magnetic Circuit MMF Chapter 6

Given:

Mean length \(l_m = 0.5\text{ m}\)
Cross-sectional area \(A = 100\text{ cm}^2 = 100 \times 10^{-4} = 0.01\text{ m}^2\)
Air gap length \(l_g = 4\text{ mm} = 0.004\text{ m}\)
Flux \(\Phi = 2\text{ mWb} = 0.002\text{ Wb}\)
Number of turns \(N = 200\)
Relative permeability \(\mu_r = 800\)
\(\mu_0 = 4\pi \times 10^{-7}\text{ H/m}\)

Calculations:

Flux density: \[B = \frac{\Phi}{A} = \frac{0.002}{0.01} = 0.2\text{ Tesla}\] Length of iron path: \[l_i = l_m - l_g = 0.5 - 0.004 = 0.496\text{ m}\] Magnetizing force in iron: \[H_i = \frac{B}{\mu_0 \mu_r} = \frac{0.2}{4\pi \times 10^{-7} \times 800} \approx 198.94\text{ AT/m}\] MMF for iron path: \[\text{MMF}_i = H_i \times l_i = 198.94 \times 0.496 \approx 98.68\text{ AT}\] Magnetizing force in air gap: \[H_g = \frac{B}{\mu_0} = \frac{0.2}{4\pi \times 10^{-7}} \approx 159,155\text{ AT/m}\] MMF for air gap: \[\text{MMF}_g = H_g \times l_g = 159,155 \times 0.004 \approx 636.62\text{ AT}\] Total MMF required: \[\text{Total MMF} = \text{MMF}_i + \text{MMF}_g = 98.68 + 636.62 = 735.30\text{ AT}\] Exciting current: \(I = \frac{\text{MMF}}{N} = \frac{735.3}{200} \approx 3.68\text{ A}\).

Final Answer: Total MMF required = \(735.3\text{ AT}\) (Exciting current = \(3.68\text{ A}\)). Notice that the tiny 4 mm air gap accounts for 86.6% of the total MMF!

Question 9 [5 Marks]: Transformer Overload Efficiency Calculation

Transformer Efficiency at 25% Overload Chapter 7

Given:

Rated capacity \(S = 150\text{ kVA}\)
Iron loss \(P_i = 700\text{ W} = 0.7\text{ kW}\) (constant)
Full-load copper loss \(P_{cu,FL} = 1800\text{ W} = 1.8\text{ kW}\)
Load condition: 25% overloaded \(\implies\) load fraction \(x = 1.25\)
Power factor \(\cos\phi = 0.8\) lagging

Calculations:

1. Output Power: \[P_{\text{out}} = x \times S \times \cos\phi = 1.25 \times 150 \times 0.8 = 150\text{ kW}\] 2. Copper Loss at \(x = 1.25\): \[P_{cu} = x^2 \times P_{cu,FL} = (1.25)^2 \times 1800 = 1.5625 \times 1800 = 2812.5\text{ W} = 2.8125\text{ kW}\] 3. Total Losses: \[\text{Total Losses} = P_i + P_{cu} = 700 + 2812.5 = 3512.5\text{ W} = 3.5125\text{ kW}\] 4. Efficiency: \[\eta = \frac{P_{\text{out}}}{P_{\text{out}} + \text{Losses}} \times 100\% = \frac{150}{150 + 3.5125} \times 100\% = \frac{150}{153.5125} \times 100\% \approx 97.71\%\]

Final Results: Copper loss = \(2812.5\text{ W}\), Iron loss = \(700\text{ W}\), Output power = \(150\text{ kW}\), Efficiency \(\eta = \mathbf{97.71\%}\).

TRIBHUVAN UNIVERSITY • IOE EXAMINATION

2082 Baishakh — Back Exam (New Course)

Program: BEI, BCT | Year/Part: I / II | Full Marks: 60 | Pass Marks: 24 | Time: 3 hrs

Question 3 [6 Marks]: Series RLC Step Response (Classical Method)

Question Statement: A DC source of 100 V is suddenly applied at time \(t = 0\) to a series RLC circuit comprising \(R = 2\,\Omega, L = 0.5\text{H}\) and \(C = 1\text{F}\). Obtain the expression for current in the circuit by using classical method.

Series RLC Classical Solution Chapter 2

Calculations:

Characteristic equation: \[L s^2 + R s + \frac{1}{C} = 0 \implies 0.5 s^2 + 2 s + 1 = 0 \implies s^2 + 4s + 2 = 0\] Roots: \[s_{1,2} = \frac{-4 \pm \sqrt{16 - 8}}{2} = -2 \pm \sqrt{2}\] Since roots are real, negative, and unequal (\(s_1 = -2 + \sqrt{2} \approx -0.586\), \(s_2 = -2 - \sqrt{2} \approx -3.414\)), the circuit is overdamped! \[i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}\] Initial conditions:
1. \(i(0^+) = 0 \implies A_1 + A_2 = 0 \implies A_2 = -A_1\).
2. Across inductor at \(t = 0^+\): \(L \frac{di}{dt}(0^+) = V \implies \frac{di}{dt}(0^+) = \frac{100}{0.5} = 200\text{ A/s}\). \[\frac{di}{dt}(0^+) = s_1 A_1 + s_2 A_2 = A_1(s_1 - s_2) = A_1[(-2+\sqrt{2}) - (-2-\sqrt{2})] = A_1(2\sqrt{2}) = 200\] \[A_1 = \frac{200}{2\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2} \approx 70.71\text{ A}\]

Final Expression: \[i(t) = 50\sqrt{2} \left[ e^{(-2+\sqrt{2})t} - e^{(-2-\sqrt{2})t} \right]\text{ A} \quad (t \ge 0)\]

Question 7 [5 Marks]: Transformer OC and SC Test Equivalent Parameters

OC & SC Test Calculations Chapter 7

Given:

20 kVA, 2200/220 V, 50 Hz transformer.
Transformation ratio referred to secondary (LV): \(K = \frac{V_2}{V_1} = \frac{220}{2200} = 0.1\).
OC test on LV: \(V_0 = 220\text{ V}, I_0 = 1.1\text{ A}, W_0 = 125\text{ W}\).
SC test on HV: \(V_{sc} = 52.7\text{ V}, I_{sc} = 8.4\text{ A}, W_{sc} = 287\text{ W}\).

Calculations:

1. From OC Test (Shunt parameters directly on LV/secondary): \[\cos\phi_0 = \frac{W_0}{V_0 I_0} = \frac{125}{220 \times 1.1} = \frac{125}{242} \approx 0.5165\] \[\sin\phi_0 = \sqrt{1 - 0.5165^2} = 0.8563\] \[I_w = I_0 \cos\phi_0 = 1.1 \times 0.5165 \approx 0.568\text{ A}\] \[I_\mu = I_0 \sin\phi_0 = 1.1 \times 0.8563 \approx 0.942\text{ A}\] \[R_0' = \frac{V_0}{I_w} = \frac{220}{0.568} \approx 387.3\,\Omega\] \[X_0' = \frac{V_0}{I_\mu} = \frac{220}{0.942} \approx 233.5\,\Omega\] 2. From SC Test (Series parameters on HV/primary): \[R_{eq1} = \frac{W_{sc}}{I_{sc}^2} = \frac{287}{8.4^2} = \frac{287}{70.56} \approx 4.067\,\Omega\] \[Z_{eq1} = \frac{V_{sc}}{I_{sc}} = \frac{52.7}{8.4} \approx 6.274\,\Omega\] \[X_{eq1} = \sqrt{Z_{eq1}^2 - R_{eq1}^2} = \sqrt{6.274^2 - 4.067^2} = \sqrt{39.36 - 16.54} = \sqrt{22.82} \approx 4.777\,\Omega\] 3. Referring Series Parameters to Secondary (LV): \[R_{eq2} = K^2 R_{eq1} = (0.1)^2 \times 4.067 = 0.04067\,\Omega\] \[X_{eq2} = K^2 X_{eq1} = (0.1)^2 \times 4.777 = 0.04777\,\Omega\] \[Z_{eq2} = K^2 Z_{eq1} = (0.1)^2 \times 6.274 = 0.06274\,\Omega\]

Parameters Referred to Secondary: \[R_{eq2} = 0.0407\,\Omega, \quad X_{eq2} = 0.0478\,\Omega, \quad R_0' = 387.3\,\Omega, \quad X_0' = 233.5\,\Omega\]

TRIBHUVAN UNIVERSITY • IOE EXAMINATION

2081 Ashwin — Regular Exam (New Course - 2080 Batch)

Program: BEI, BCT | Year/Part: I / II | Full Marks: 60 | Pass Marks: 24 | Time: 3 hrs

Question 8 [6 Marks]: DC Shunt Motor Armature Speed Control Numerical

DC Shunt Motor Speed with Added Resistance Chapter 8

Given:

Supply voltage \(V = 240\text{ V}\)
Armature resistance \(R_a = 0.4\,\Omega\)
Field resistance \(R_{sh} = 120\,\Omega\)
Half-load line current \(I_{L1} = 30\text{ A}\)
Initial speed \(N_1 = 1400\text{ rpm}\)
Added armature series resistance \(R_{\text{ext}} = 1.2\,\Omega\)
Load torque is reduced by 20% \(\implies T_2 = 0.8 T_1\)

Calculations:

1. Shunt field current (constant): \[I_{sh} = \frac{V}{R_{sh}} = \frac{240}{120} = 2\text{ A}\] 2. Initial armature current: \[I_{a1} = I_{L1} - I_{sh} = 30 - 2 = 28\text{ A}\] 3. Initial back EMF: \[E_{b1} = V - I_{a1} R_a = 240 - (28 \times 0.4) = 240 - 11.2 = 228.8\text{ V}\] 4. Torque relationship (field current is constant \(\implies \Phi_1 = \Phi_2\)): \[\frac{T_2}{T_1} = \frac{I_{a2}}{I_{a1}} = 0.8 \implies I_{a2} = 0.8 \times 28 = 22.4\text{ A}\] 5. New total armature circuit resistance: \[R_{a2} = R_a + R_{\text{ext}} = 0.4 + 1.2 = 1.6\,\Omega\] 6. New back EMF: \[E_{b2} = V - I_{a2} R_{a2} = 240 - (22.4 \times 1.6) = 240 - 35.84 = 204.16\text{ V}\] 7. Speed calculation (\(N \propto E_b / \Phi\)): \[\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} = \frac{204.16}{228.8} \approx 0.8923\] \[N_2 = 1400 \times 0.8923 \approx 1249.2\text{ rpm}\]

Final Answer: The new motor speed is \(1249.2\text{ rpm}\).