Exam Derivation Master Sheet

Master Derivations for ENEE 154

Rigorous, step-by-step mathematical proofs frequently asked in IOE final exams. Formatted with stated starting principles, explicit assumptions, and physical significance.

Chapter 9 • AC Motors 2082 Baishakh Q9 • 2081 Ashwin Q9

1. Production of Rotating Magnetic Field (RMF) in 3-Phase Induction Motor

Starting Principle: When three-phase balanced stator windings displaced in space by \(120^\circ\) electrical are energized by balanced three-phase currents displaced in time by \(120^\circ\), they produce a resultant rotating magnetic flux.

Mathematical Formulation: Let the three pulsating magnetic fluxes produced by phases R, Y, and B along their respective spatial axes be:

\[\phi_R = \Phi_m \cos(\omega t)\] \[\phi_Y = \Phi_m \cos(\omega t - 120^\circ)\] \[\phi_B = \Phi_m \cos(\omega t - 240^\circ) = \Phi_m \cos(\omega t + 120^\circ)\]

Let the spatial axis of phase R be along the horizontal reference (\(\theta = 0^\circ\)). The spatial axes of Y and B are at \(-120^\circ\) and \(-240^\circ\) (or \(+120^\circ\)).

Step-by-step Resolution along Orthogonal Axes:

Horizontal Component (\(\Phi_x\)):

\[\Phi_x = \phi_R \cos(0^\circ) + \phi_Y \cos(-120^\circ) + \phi_B \cos(120^\circ)\] \[\Phi_x = \phi_R(1) + \phi_Y \left(-\frac{1}{2}\right) + \phi_B \left(-\frac{1}{2}\right) = \phi_R - \frac{1}{2}(\phi_Y + \phi_B)\] Since the currents are balanced, \(\phi_R + \phi_Y + \phi_B = 0 \implies \phi_Y + \phi_B = -\phi_R\). Substituting: \[\Phi_x = \phi_R - \frac{1}{2}(-\phi_R) = \frac{3}{2}\phi_R = \frac{3}{2}\Phi_m \cos(\omega t)\]

Vertical Component (\(\Phi_y\)):

\[\Phi_y = \phi_R \sin(0^\circ) + \phi_Y \sin(-120^\circ) + \phi_B \sin(120^\circ)\] \[\Phi_y = 0 + \phi_Y \left(-\frac{\sqrt{3}}{2}\right) + \phi_B \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}(\phi_B - \phi_Y)\] Substitute the expressions for \(\phi_B\) and \(\phi_Y\): \[\phi_B - \phi_Y = \Phi_m [\cos(\omega t + 120^\circ) - \cos(\omega t - 120^\circ)]\] Using identity \(\cos(A+B) - \cos(A-B) = -2\sin A \sin B\): \[\phi_B - \phi_Y = \Phi_m [-2\sin(\omega t)\sin(120^\circ)] = -2\Phi_m \sin(\omega t)\left(\frac{\sqrt{3}}{2}\right) = -\sqrt{3}\Phi_m \sin(\omega t)\] Now substitute back into \(\Phi_y\): \[\Phi_y = \frac{\sqrt{3}}{2} [-\sqrt{3}\Phi_m \sin(\omega t)] = -\frac{3}{2}\Phi_m \sin(\omega t)\]

Resultant Flux Magnitude (\(\Phi_r\)):

\[\Phi_r = \sqrt{\Phi_x^2 + \Phi_y^2} = \sqrt{\left(\frac{3}{2}\Phi_m \cos(\omega t)\right)^2 + \left(-\frac{3}{2}\Phi_m \sin(\omega t)\right)^2} = \frac{3}{2}\Phi_m \sqrt{\cos^2(\omega t) + \sin^2(\omega t)} = \frac{3}{2}\Phi_m = 1.5\Phi_m\]

Orientation Angle (\(\theta\)):

\[\tan\theta = \frac{\Phi_y}{\Phi_x} = \frac{-\frac{3}{2}\Phi_m \sin(\omega t)}{\frac{3}{2}\Phi_m \cos(\omega t)} = -\tan(\omega t) \implies \theta = -\omega t\]
Physical Significance: The resultant magnetic field has a constant magnitude equal to \(1.5 \Phi_m\) (1.5 times the maximum pulsating flux of any individual phase) and rotates smoothly in the air gap at constant synchronous speed \(N_s = \frac{120 f}{P}\) rpm.
Chapter 7 • Transformers 2083 Baishakh Q7 • 2082 Bhadra Q10b • 2081 Ashwin Q7

2. Proof that Main Magnetic Core Flux is Independent of Load Current

Starting Principle: Faraday's Law of Induction and primary voltage balance.

In a practical transformer, the primary induced back EMF \(E_1\) is given by:

\[V_1 = -E_1 + I_1 Z_1 \implies E_1 \approx V_1 \quad (\text{since primary leakage impedance drop } I_1 Z_1 \text{ is negligibly small, } < 2-3\%)\] The RMS value of induced EMF is: \[E_1 = 4.44 f N_1 \Phi_m\] Equating \(E_1 \approx V_1\): \[\Phi_m \approx \frac{V_1}{4.44 f N_1}\] Since supply voltage \(V_1\) and frequency \(f\) are maintained constant by the AC grid, \(\Phi_m\) must remain practically constant regardless of load current!

Physical Mechanism (MMF Balance):

  1. At no load, primary draws a small no-load current \(I_0\) which provides magnetizing MMF \(N_1 I_0\) to establish the mutual core flux \(\Phi_m\).
  2. When secondary is loaded with current \(I_2\), it creates a demagnetizing MMF \(N_2 I_2\) setting up opposing flux \(\Phi_2\).
  3. This momentarily tends to reduce core flux \(\Phi_m\), which instantly causes back EMF \(E_1\) to drop slightly.
  4. Because \(V_1\) is constant, the small drop in \(E_1\) allows an additional current \(I_1'\) (load component of primary current) to be drawn from the supply such that: \[N_1 I_1' = N_2 I_2 \implies I_1' = \left(\frac{N_2}{N_1}\right) I_2 = K I_2\]
  5. The MMF \(N_1 I_1'\) exactly neutralizes the demagnetizing MMF \(N_2 I_2\) at every instant!
\[\text{Net Core MMF} = N_1 I_1 - N_2 I_2 = N_1(I_0 + I_1') - N_2 I_2 = N_1 I_0 = \text{Constant}\]
Conclusion: The core flux \(\Phi_m\) is strictly established by the primary voltage \(V_1\) and frequency \(f\), and remains completely constant from no-load to full-load.
Chapter 8 • DC Machines 2082 Baishakh Q8

3. Derivation of DC Motor Torque Equation

Starting Principle: Conservation of electrical to mechanical power conversion in the armature.

Let:

  • \(P\) = Number of field poles
  • \(\Phi\) = Magnetic flux per pole [Wb]
  • \(Z\) = Total number of armature conductors
  • \(A\) = Number of parallel armature paths (\(A = P\) for lap, \(A = 2\) for wave)
  • \(N\) = Armature speed in rpm (\(\omega = \frac{2\pi N}{60}\) rad/s)
  • \(I_a\) = Total armature current drawn from supply [A]
  • \(E_b\) = Back EMF induced in the armature [V]

Step 1: Generated Back EMF

\[E_b = \frac{P \Phi Z N}{60 A}\]

Step 2: Power Conversion Balance

The total electrical power converted into gross mechanical power in the armature is:

\[P_{\text{mech}} = E_b I_a\] Mechanical power can also be expressed in terms of electromagnetic torque \(T\) and angular velocity \(\omega\): \[P_{\text{mech}} = T \omega = T \left(\frac{2\pi N}{60}\right)\]

Step 3: Equating Powers and Solving for \(T\)

\[T \left(\frac{2\pi N}{60}\right) = E_b I_a = \left(\frac{P \Phi Z N}{60 A}\right) I_a\] Canceling \(\frac{N}{60}\) from both sides: \[T (2\pi) = \frac{P \Phi Z I_a}{A} \implies T = \frac{P \Phi Z I_a}{2\pi A} \text{ [N}\cdot\text{m]}\] \[T = \left(\frac{1}{2\pi}\right) \frac{P Z}{A} \Phi I_a \approx 0.159 \frac{P Z}{A} \Phi I_a \implies T \propto \Phi I_a\]
Key Result: Electromagnetic torque is directly proportional to the product of field flux \(\Phi\) and armature current \(I_a\).
Chapter 2 • Classical Transients 2082 Bhadra Q3 • 2082 Baishakh Q3

4. Series R-L-C Circuit Step Response Differential Equation & Solutions

Circuit: Series connection of \(R\), \(L\), and \(C\) connected to DC step voltage source \(V\) switched at \(t=0\).

Applying KVL for \(t > 0\):

\[R i(t) + L \frac{di(t)}{dt} + \frac{1}{C} \int_0^t i(\tau) d\tau + v_C(0^+) = V\] Differentiating both sides with respect to \(t\): \[L \frac{d^2 i}{dt^2} + R \frac{di}{dt} + \frac{1}{C} i = 0 \implies \frac{d^2 i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{1}{LC} i = 0\] This is a second-order linear homogeneous ordinary differential equation with constant coefficients.

Characteristic Equation: Substituting \(i(t) = A e^{st}\):

\[s^2 + \frac{R}{L} s + \frac{1}{LC} = 0 \implies s^2 + 2\alpha s + \omega_0^2 = 0\] Where: \[\alpha = \frac{R}{2L} \text{ (Neper damping frequency [Np/s])}, \quad \omega_0 = \frac{1}{\sqrt{LC}} \text{ (Undamped resonant frequency [rad/s])}\] Roots: \[s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}\]

Three Cases of Circuit Response:

Condition Roots Nature Current Solution \(i(t)\) Physical Behavior
Case 1: Overdamped (\(\alpha > \omega_0 \implies R > 2\sqrt{L/C}\)) Real, negative, unequal: \(s_1 \ne s_2 < 0\) \(i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}\) Sluggish exponential rise and decay without any oscillations.
Case 2: Critically Damped (\(\alpha = \omega_0 \implies R = 2\sqrt{L/C}\)) Real, negative, equal: \(s_1 = s_2 = -\alpha\) \(i(t) = (A_1 + A_2 t) e^{-\alpha t}\) Fastest non-oscillatory return to equilibrium. Ideal for meter movements.
Case 3: Underdamped (\(\alpha < \omega_0 \implies R < 2\sqrt{L/C}\)) Complex conjugate: \(-\alpha \pm j\omega_d\), \(\omega_d = \sqrt{\omega_0^2 - \alpha^2}\) \(i(t) = e^{-\alpha t}[A_1 \cos(\omega_d t) + A_2 \sin(\omega_d t)]\) Damped sinusoidal oscillations at damped frequency \(\omega_d\).
Chapter 2 • Classical Transients 2083 Baishakh Q3

5. Series R-C Circuit with Exponential Voltage Source \(v(t) = V_0 e^{-at}\)

Governing KVL Equation:

\[R i(t) + \frac{1}{C}\int i(t) dt = V_0 e^{-at}\] Differentiating with respect to \(t\): \[R \frac{di}{dt} + \frac{1}{C} i = -a V_0 e^{-at} \implies \frac{di}{dt} + \frac{1}{RC} i = -\frac{a V_0}{R} e^{-at}\]

Step 1: Complementary Function (\(i_c(t)\)):

\[\frac{di}{dt} + \frac{1}{RC} i = 0 \implies i_c(t) = A e^{-t/RC}\]

Step 2: Particular Integral (\(i_p(t)\)) by Undetermined Coefficients:

Assuming \(a \ne \frac{1}{RC}\), trial solution is \(i_p(t) = K e^{-at}\). Substitute into ODE:

\[-a K e^{-at} + \frac{1}{RC} K e^{-at} = -\frac{a V_0}{R} e^{-at}\] \[K \left(\frac{1}{RC} - a\right) = -\frac{a V_0}{R} \implies K = \frac{-a V_0 / R}{\frac{1}{RC} - a} = \frac{-a C V_0}{1 - a RC} = \frac{a C V_0}{a RC - 1}\] Therefore, complete general solution is: \[i(t) = i_c(t) + i_p(t) = A e^{-t/RC} + \frac{a C V_0}{a RC - 1} e^{-at}\]

Step 3: Evaluating Constant \(A\) from Initial Conditions:

Assuming uncharged capacitor at \(t = 0^-\), \(v_C(0^+) = v_C(0^-) = 0\). At \(t = 0^+\), across resistor:

\[R i(0^+) + v_C(0^+) = v(0^+) \implies R i(0^+) + 0 = V_0 \implies i(0^+) = \frac{V_0}{R}\] Substitute \(t = 0\) into general solution: \[i(0^+) = A + \frac{a C V_0}{a RC - 1} = \frac{V_0}{R} \implies A = \frac{V_0}{R} - \frac{a C V_0}{a RC - 1} = \frac{V_0(a RC - 1) - a R C V_0}{R(a RC - 1)} = \frac{-V_0}{R(a RC - 1)}\] Substituting \(A\) back into \(i(t)\): \[i(t) = \frac{V_0}{R(1 - a RC)} \left[ e^{-t/RC} - a RC \, e^{-at} \right]\]
Chapter 9 • AC Motors 2081 Ashwin Q9

6. Condition for Maximum Running Torque in 3-Phase Induction Motor

Torque Expression: Rotor running torque is:

\[T = \frac{k \, s E_2^2 R_2}{R_2^2 + (s X_2)^2}\] To find slip \(s\) at which torque is maximum, differentiate \(T\) with respect to \(s\) and set to zero: \[\frac{dT}{ds} = k E_2^2 R_2 \left[ \frac{(R_2^2 + s^2 X_2^2)(1) - s(2s X_2^2)}{(R_2^2 + s^2 X_2^2)^2} \right] = 0\] \[R_2^2 + s^2 X_2^2 - 2s^2 X_2^2 = 0 \implies R_2^2 - s^2 X_2^2 = 0 \implies s^2 X_2^2 = R_2^2\] \[s_{mT} = \frac{R_2}{X_2}\]

Maximum Torque Expression: Substitute \(s = s_{mT} = R_2 / X_2\) into the torque equation:

\[T_{\text{max}} = \frac{k \left(\frac{R_2}{X_2}\right) E_2^2 R_2}{R_2^2 + \left(\frac{R_2}{X_2}\right)^2 X_2^2} = \frac{k E_2^2 \left(\frac{R_2^2}{X_2}\right)}{R_2^2 + R_2^2} = \frac{k E_2^2 \left(\frac{R_2^2}{X_2}\right)}{2 R_2^2} = \frac{k E_2^2}{2 X_2}\]
Vital Deductions:
1. Slip at max torque is directly proportional to rotor resistance: \(s_{mT} \propto R_2\).
2. Peak maximum torque \(T_{\text{max}}\) is independent of rotor resistance \(R_2\)!
3. Adding external resistance to a wound rotor shifts the peak torque toward standstill (\(s=1\)), giving higher starting torque without decreasing breakdown torque.
Chapter 5 • Two-Port Networks 2082 Baishakh Q6

7. Equivalent Transmission (ABCD) Parameters of Two Cascade-Connected Networks

Let network \(N_a\) be followed by network \(N_b\). The output of \(N_a\) is connected to the input of \(N_b\).

Terminal Conditions at Interconnection:

\[V_{2a} = V_{1b}, \quad -I_{2a} = I_{1b} \implies I_{2a} = -I_{1b}\]

Matrix Formulation:

For Network A: \[\begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} A_a & B_a \\ C_a & D_a \end{bmatrix} \begin{bmatrix} V_{2a} \\ -I_{2a} \end{bmatrix}\] Substitute terminal conditions: \[\begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} A_a & B_a \\ C_a & D_a \end{bmatrix} \begin{bmatrix} V_{1b} \\ I_{1b} \end{bmatrix}\] For Network B: \[\begin{bmatrix} V_{1b} \\ I_{1b} \end{bmatrix} = \begin{bmatrix} A_b & B_b \\ C_b & D_b \end{bmatrix} \begin{bmatrix} V_2 \\ -I_2 \end{bmatrix}\] Substituting the equation of Network B into Network A: \[\begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} A_a & B_a \\ C_a & D_a \end{bmatrix} \begin{bmatrix} A_b & B_b \\ C_b & D_b \end{bmatrix} \begin{bmatrix} V_2 \\ -I_2 \end{bmatrix}\] Thus, the overall transmission matrix is the direct matrix product: \[\begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A_a A_b + B_a C_b & A_a B_b + B_a D_b \\ C_a A_b + D_a C_b & C_a B_b + D_a D_b \end{bmatrix}\]
Chapter 5 • Two-Port Networks 2082 Baishakh Q6

8. Transformation of Hybrid (\(h\)) Parameters from Impedance (\(Z\)) Parameters

Starting Equations:

Z-parameter equations: \[(1) \quad V_1 = Z_{11} I_1 + Z_{12} I_2\] \[(2) \quad V_2 = Z_{21} I_1 + Z_{22} I_2\] Target h-parameter equations: \[(3) \quad V_1 = h_{11} I_1 + h_{12} V_2\] \[(4) \quad I_2 = h_{21} I_1 + h_{22} V_2\]

Step 1: Express \(I_2\) in terms of \(I_1\) and \(V_2\)

From equation (2): \[Z_{22} I_2 = V_2 - Z_{21} I_1 \implies I_2 = -\frac{Z_{21}}{Z_{22}} I_1 + \frac{1}{Z_{22}} V_2\] Comparing with target equation (4): \[h_{21} = -\frac{Z_{21}}{Z_{22}}, \quad h_{22} = \frac{1}{Z_{22}}\]

Step 2: Substitute \(I_2\) into equation (1) for \(V_1\)

\[V_1 = Z_{11} I_1 + Z_{12} \left( -\frac{Z_{21}}{Z_{22}} I_1 + \frac{1}{Z_{22}} V_2 \right) = \left( Z_{11} - \frac{Z_{12} Z_{21}}{Z_{22}} \right) I_1 + \frac{Z_{12}}{Z_{22}} V_2\] \[V_1 = \left( \frac{Z_{11} Z_{22} - Z_{12} Z_{21}}{Z_{22}} \right) I_1 + \frac{Z_{12}}{Z_{22}} V_2 = \frac{\Delta Z}{Z_{22}} I_1 + \frac{Z_{12}}{Z_{22}} V_2\] Comparing with target equation (3): \[h_{11} = \frac{\Delta Z}{Z_{22}}, \quad h_{12} = \frac{Z_{12}}{Z_{22}}\]
Chapter 7 • Transformers 2082 Baishakh Q10b

9. Saving of Copper in an Auto-Transformer compared to 2-Winding Transformer

Weight of conductor material is proportional to volume = length \(\times\) area \(\propto N \times I\).

For a standard 2-winding transformer:

\[W_{\text{2-wdg}} \propto N_1 I_1 + N_2 I_2\] Since \(N_1 I_1 = N_2 I_2\): \[W_{\text{2-wdg}} \propto 2 N_1 I_1\] For a step-down auto-transformer with total turns \(N_1\) and tapped turns \(N_2\):
  • Top section (turns \(N_1 - N_2\)) carries current \(I_1\).
  • Common section (turns \(N_2\)) carries current \((I_2 - I_1)\).
\[W_{\text{auto}} \propto (N_1 - N_2) I_1 + N_2 (I_2 - I_1) = N_1 I_1 - N_2 I_1 + N_2 I_2 - N_2 I_1 = N_1 I_1 + N_2 I_2 - 2 N_2 I_1\] Since \(N_2 I_2 = N_1 I_1\) and \(\frac{N_2}{N_1} = K\): \[W_{\text{auto}} \propto 2 N_1 I_1 - 2 K N_1 I_1 = 2 N_1 I_1 (1 - K)\] Taking the ratio: \[\frac{W_{\text{auto}}}{W_{\text{2-wdg}}} = \frac{2 N_1 I_1 (1 - K)}{2 N_1 I_1} = 1 - K\] \[\text{Saving of Copper} = \frac{W_{\text{2-wdg}} - W_{\text{auto}}}{W_{\text{2-wdg}}} = 1 - (1 - K) = K\]
Conclusion: When the transformation ratio \(K\) is close to 1 (e.g., \(K = 0.9\)), the autotransformer saves up to 90% of copper compared to a conventional two-winding transformer.