BE Computer Engineering (IOE, TU) Basic Electrical Engineering (IOE, EE 451) Question Paper 2079 Nepal

(a) Thevenin's Theorem

Statement: Any linear, bilateral two-terminal network containing sources and resistances can be replaced, with respect to two terminals A-B, by a single voltage source $V_{Th}$ in series with a single resistance $R_{Th}$, where $V_{Th}$ is the open-circuit voltage across A-B and $R_{Th}$ is the equivalent resistance seen from A-B with all independent sources deactivated.

Procedure (network with independent and dependent sources):

1. Remove the load $R_L$ from terminals A-B.
2. Find the open-circuit voltage $V_{Th} = V_{OC}$ across A-B (mesh/nodal analysis).
3. To find $R_{Th}$:
   • If only independent sources: deactivate them (replace voltage sources by short circuits, current sources by open circuits) and compute the resistance looking into A-B.
   • If dependent sources are present: deactivate only the independent sources, apply a test source $V_t$ (or $I_t$) at A-B, and compute $R_{Th} = V_t / I_t$. Alternatively, $R_{Th} = V_{OC} / I_{SC}$, where $I_{SC}$ is the short-circuit current.
4. Draw the Thevenin equivalent: $V_{Th}$ in series with $R_{Th}$, then reconnect $R_L$.

(b) Numerical

With the load removed at node C (open circuit), the 24 V source drives current through $4\,\Omega + 6\,\Omega$ in series, then through the $12\,\Omega$ to ground.

Open-circuit voltage (voltage divider across the $12\,\Omega$):

$$V_{Th} = 24 \times \frac{12}{4 + 6 + 12} = 24 \times \frac{12}{22} = 13.09\ \text{V}$$

Thevenin resistance (24 V source shorted): looking into C-ground, the $12\,\Omega$ is in parallel with $(4+6)=10\,\Omega$:

$$R_{Th} = \frac{12 \times 10}{12 + 10} = \frac{120}{22} = 5.45\ \Omega$$

Maximum power transfer: $R_L = R_{Th} = 5.45\ \Omega$.

$$P_{max} = \frac{V_{Th}^2}{4R_{Th}} = \frac{(13.09)^2}{4 \times 5.45} = \frac{171.4}{21.8} = 7.86\ \text{W}$$

(a) EMF Equation of a Single-Phase Transformer

When a sinusoidal flux $\phi = \phi_m \sin\omega t$ links a winding of $N$ turns, the induced EMF is $e = -N\,d\phi/dt$. The flux rises from 0 to $\phi_m$ in a quarter cycle $(1/4f)$ seconds, so the average rate of change is $\frac{\phi_m}{1/4f} = 4f\phi_m$, giving average EMF per turn $= 4f\phi_m$.

For a sinusoid, RMS = form factor (1.11) × average:

$$E = 1.11 \times 4f\phi_m N = 4.44\,f\,N\,\phi_m$$

Thus $E_1 = 4.44 f N_1 \phi_m$ and $E_2 = 4.44 f N_2 \phi_m$.

Transformation ratio:

$$K = \frac{E_2}{E_1} = \frac{N_2}{N_1} = \frac{V_2}{V_1} = \frac{I_1}{I_2}$$

(b) Numerical (referred to HV side)

Rated HV current $I_2 = \dfrac{25000}{2200} = 11.36\ \text{A}$.

From SC test (HV side): $V_{sc}=75\,$V, $I_{sc}=11.4\,$A, $W_{sc}=280\,$W

$$Z_{eq} = \frac{75}{11.4} = 6.58\ \Omega,\quad R_{eq} = \frac{280}{11.4^2} = 2.15\ \Omega,\quad X_{eq} = \sqrt{6.58^2 - 2.15^2} = 6.22\ \Omega$$

Equivalent circuit (HV side): series $R_{eq}=2.15\,\Omega$ and $X_{eq}=6.22\,\Omega$ in the line; shunt magnetising branch ($R_0, X_0$) across the input, with core loss $P_i = 90$ W from the OC test.

Losses at full load:

• Core (iron) loss $P_i = 90\ \text{W}$ (OC test).
• Full-load copper loss $P_{cu} = I_2^2 R_{eq} = 11.36^2 \times 2.15 = 277.5\ \text{W}$.

Efficiency at full load, 0.8 pf:

$$\eta = \frac{25000 \times 0.8}{25000 \times 0.8 + 90 + 277.5} = \frac{20000}{20367.5} = 98.2\%$$

Voltage regulation (0.8 pf lagging, $\cos\phi=0.8,\ \sin\phi=0.6$):

$$\%\text{Reg} = \frac{I_2(R_{eq}\cos\phi + X_{eq}\sin\phi)}{V_2}\times100 = \frac{11.36(2.15\times0.8 + 6.22\times0.6)}{2200}\times100$$ $$= \frac{11.36 \times 5.45}{2200}\times100 = 2.82\%$$

(a) DC Shunt Generator

Construction: Stationary part (yoke, poles with field windings) and rotating armature carrying conductors in slots, with a commutator and brushes. In a shunt generator the field winding (many turns of fine wire) is connected in parallel with the armature.

Working principle: Based on Faraday's law of electromagnetic induction. When the armature rotates in the magnetic field, the conductors cut flux and an EMF is induced. The commutator and brushes convert the internally generated AC into DC at the terminals. (Sketch: poles N-S on the stator, armature conductors between them, commutator + brushes feeding the field and load.)

EMF equation: Let $P$ = poles, $Z$ = total armature conductors, $\phi$ = flux/pole, $N$ = speed (rpm), $A$ = parallel paths. Flux cut per conductor per revolution $= P\phi$; revolutions per second $= N/60$. EMF per conductor $= P\phi N/60$. With $Z/A$ conductors in series per path:

$$E_g = \frac{P\phi Z N}{60 A}\ \text{volts}$$

($A = P$ for lap, $A = 2$ for wave winding.)

(b) DC Shunt Motor Numerical

Field current $I_{sh} = \dfrac{220}{110} = 2\ \text{A}$. Armature current (full load) $I_a = I_L - I_{sh} = 40 - 2 = 38\ \text{A}$.

Back EMF at full load:

$$E_{b1} = V - I_a R_a = 220 - 38 \times 0.25 = 220 - 9.5 = 210.5\ \text{V}$$

When $I_a$ falls to 20 A:

$$E_{b2} = 220 - 20 \times 0.25 = 215\ \text{V}$$

Since $E_b \propto N\phi$ and flux is constant, $\dfrac{N_2}{N_1} = \dfrac{E_{b2}}{E_{b1}}$:

$$N_2 = 1200 \times \frac{215}{210.5} = 1225.7\ \text{rpm}$$

(a) Series RLC Resonance

Definition: A series RLC circuit is in resonance when the inductive reactance equals the capacitive reactance ($X_L = X_C$); the net reactance is zero, the impedance is minimum ($Z = R$), the current is maximum, and the circuit behaves purely resistively (current in phase with voltage).

Resonant frequency: Setting $\omega_0 L = 1/\omega_0 C$:

$$\omega_0 = \frac{1}{\sqrt{LC}}\quad\Rightarrow\quad f_0 = \frac{1}{2\pi\sqrt{LC}}$$

Quality factor: ratio of reactive to resistive voltage at resonance:

$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R}\sqrt{\frac{L}{C}}$$

Bandwidth (between half-power frequencies $f_1, f_2$):

$$BW = f_2 - f_1 = \frac{R}{2\pi L} = \frac{f_0}{Q}$$

Current vs frequency: the current $I = V/Z$ is a bell-shaped curve, zero at very low and very high frequency, rising to a sharp peak $I_{max}=V/R$ at $f_0$; the sharper the peak, the higher the Q.

(b) Numerical ($R=10\,\Omega,\ L=0.1\,$H, $C=100\,\mu$F, $V=230\,$V)

(i) Resonant frequency:

$$f_0 = \frac{1}{2\pi\sqrt{0.1 \times 100\times10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-5}}} = 50.3\ \text{Hz}$$

(ii) Impedance at resonance: $Z = R = 10\ \Omega$.

(iii) Current at resonance: $I = V/R = 230/10 = 23\ \text{A}$.

(iv) Quality factor and bandwidth:

$$Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{0.1}{100\times10^{-6}}} = \frac{1}{10}\sqrt{1000} = 3.16$$ $$BW = \frac{f_0}{Q} = \frac{50.3}{3.16} = 15.9\ \text{Hz}$$

Balanced Star-Connected Load

Per-phase impedance $Z = 8 + j6\ \Omega$, so $|Z| = \sqrt{8^2+6^2} = 10\ \Omega$, angle $\phi = \tan^{-1}(6/8) = 36.87^\circ$.

Phase voltage (star): $V_{ph} = \dfrac{V_L}{\sqrt3} = \dfrac{400}{\sqrt3} = 231\ \text{V}$.

Phase current $= \dfrac{V_{ph}}{|Z|} = \dfrac{231}{10} = 23.1\ \text{A}$.

Line current (star: $I_L = I_{ph}$) $= 23.1\ \text{A}$.

Power factor $= \cos 36.87^\circ = \dfrac{R}{|Z|} = \dfrac{8}{10} = 0.8$ (lagging).

Total active power:

$$P = \sqrt3\,V_L I_L \cos\phi = \sqrt3 \times 400 \times 23.1 \times 0.8 = 12.8\ \text{kW}$$

Total reactive power:

$$Q = \sqrt3\,V_L I_L \sin\phi = \sqrt3 \times 400 \times 23.1 \times 0.6 = 9.6\ \text{kVAR}$$

Composite Magnetic Circuit (Iron + Air Gap)

Given: mean circumference $= 60\,$cm, $A = 5\,\text{cm}^2 = 5\times10^{-4}\,\text{m}^2$, air gap $l_g = 2\,$mm, $N=500$, $\mu_r = 800$, $\phi = 0.5\,$mWb.

Flux density (same in iron and gap, neglecting fringing):

$$B = \frac{\phi}{A} = \frac{0.5\times10^{-3}}{5\times10^{-4}} = 1.0\ \text{T}$$

MMF for iron path ($l_i = 0.60 - 0.002 = 0.598\,$m):

$$H_i = \frac{B}{\mu_0\mu_r} = \frac{1.0}{4\pi\times10^{-7}\times800} = 995\ \text{A/m}$$ $$\text{MMF}_i = H_i\,l_i = 995 \times 0.598 = 595\ \text{At}$$

MMF for air gap:

$$H_g = \frac{B}{\mu_0} = \frac{1.0}{4\pi\times10^{-7}} = 7.96\times10^{5}\ \text{A/m}$$ $$\text{MMF}_g = H_g\,l_g = 7.96\times10^{5}\times0.002 = 1592\ \text{At}$$

Total MMF $= 595 + 1592 = 2187\ \text{At}$.

Required current:

$$I = \frac{\text{Total MMF}}{N} = \frac{2187}{500} = 4.37\ \text{A}$$

Superposition Theorem

Statement: In any linear bilateral network containing more than one independent source, the response (current or voltage) in any element equals the algebraic sum of the responses produced by each independent source acting alone, with all other independent sources deactivated (voltage sources short-circuited, current sources open-circuited; internal resistances retained).

Limitations:

• Applies only to linear circuits; not valid for non-linear elements (diodes, etc.).
• Applicable only to current and voltage (linear responses), not directly to power.
• Dependent (controlled) sources are not deactivated — they remain active.

Why it cannot be applied directly to power: Power is a non-linear (quadratic) function of current/voltage, $P = I^2 R$. If a resistor carries $I_1$ from source 1 and $I_2$ from source 2, the total current is $I = I_1 + I_2$, but

$$P = (I_1 + I_2)^2 R = I_1^2 R + I_2^2 R + 2I_1 I_2 R \neq I_1^2 R + I_2^2 R.$$

The cross term $2I_1I_2R$ is lost if powers are simply added. Hence superposition is used to find the net current/voltage first, and power is then computed from that resultant.

PMMC vs Moving-Iron (MI) Instruments

Why PMMC cannot measure AC directly: Its deflecting torque is proportional to the instantaneous current and reverses direction with the current. On AC, the average torque over a cycle is zero, so the pointer merely vibrates around zero and reads nothing. (AC can be measured only with a rectifier or thermocouple ahead of the PMMC movement.)

Range extension for large currents: A low-value shunt resistance is connected in parallel with the meter coil so that most of the current bypasses the delicate coil, and only a fixed small fraction flows through it. For multiplier ratio $m = I/I_m$, the shunt is $R_{sh} = \dfrac{R_m}{m - 1}$, where $R_m$ is the coil resistance.

3-Phase Induction Motor

Given: $P = 4$ poles, $f = 50\,$Hz, $N = 1440\,$rpm.

Synchronous speed:

$$N_s = \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500\ \text{rpm}$$

Slip:

$$s = \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = 0.04 = 4\%$$

Rotor EMF frequency:

$$f_r = s f = 0.04 \times 50 = 2\ \text{Hz}$$

Why it cannot run at synchronous speed: At synchronous speed the rotor would move exactly with the rotating stator field, so there would be no relative motion, no flux cutting, no rotor EMF, no rotor current and hence no torque. The motor would decelerate. A small slip is therefore always needed to maintain the rotor current and torque; for this reason the induction motor is also called an asynchronous motor.

(a) Average, RMS and Form Factor

• Average value: the mean of the instantaneous values over half a cycle (for a symmetrical wave); $V_{av} = \dfrac{1}{T}\int_0^T v\,dt$ over the relevant interval.
• RMS value: the equivalent DC value that produces the same heating effect; $V_{rms} = \sqrt{\dfrac{1}{T}\int_0^T v^2\,dt}$.
• Form factor: $k_f = \dfrac{V_{rms}}{V_{av}}$.

Sinusoid $v = V_m\sin\theta$:

$$V_{av} = \frac{2V_m}{\pi} = 0.637\,V_m,\qquad V_{rms} = \frac{V_m}{\sqrt2} = 0.707\,V_m$$ $$k_f = \frac{0.707\,V_m}{0.637\,V_m} = 1.11$$

(b) Power Factor and Its Improvement

Power factor is the cosine of the phase angle $\phi$ between voltage and current, $\cos\phi = P/S$ (active power / apparent power). A low (lagging) pf means large reactive current, higher line losses, larger kVA demand and poor voltage regulation.

Two practical improvement methods:

1. Static capacitor bank connected in parallel with the inductive load: the leading capacitive current cancels part of the lagging reactive current, raising the pf.
2. Synchronous condenser (an over-excited synchronous motor running on no load): it draws leading reactive power and improves pf, useful for large bulk loads. (Phase-advancers on wound-rotor induction motors are another accepted method.)

(a) Two-Winding Transformer vs Auto-Transformer

• Two-winding transformer: primary and secondary are electrically isolated; energy is transferred entirely by magnetic (mutual) induction through separate windings.
• Auto-transformer: has a single tapped winding common to both sides; part of the winding is shared, so energy is transferred by both magnetic induction and direct electrical (conductive) connection.

Advantages of auto-transformer: less copper and iron (cheaper, lighter, smaller); lower copper and iron losses, hence higher efficiency; better voltage regulation; useful for small voltage ratios (starters, voltage regulation).

Limitation: no electrical isolation between primary and secondary — a fault/break in the common winding can place full primary voltage on the secondary, which is unsafe for large ratios.

(b) Hysteresis Loss vs Eddy-Current Loss

Both together constitute the core (iron) loss, which is essentially constant at constant voltage and frequency.