BE Computer Engineering (IOE, TU) Basic Electrical Engineering (IOE, EE 451) Question Paper 2078 Nepal

(a) Thevenin's Theorem

Statement: Any linear, bilateral two-terminal network containing sources and resistances can be replaced, with respect to its output terminals, by a single voltage source $V_{TH}$ in series with a single resistance $R_{TH}$.

• $V_{TH}$ = open-circuit voltage measured across the two terminals (load removed).
• $R_{TH}$ = equivalent resistance seen from the terminals when all independent voltage sources are short-circuited and current sources open-circuited (load removed).

The theorem simplifies the analysis of the current through a single variable load.

(b) Numerical Solution

The load is the 5 Ω resistor placed across the parallel combination (12 Ω $\parallel$ 4 Ω). Remove it.

Step 1 — Source-side series resistance: internal resistance + 6 Ω series $= 2 + 6 = 8\ \Omega$.

Step 2 — Thevenin voltage (open-circuit voltage across the parallel pair):

$$12\parallel4 = \frac{12\times4}{16}=3\ \Omega$$ $$V_{TH}=20\times\frac{3}{8+3}=\frac{60}{11}=5.45\ \text{V}$$

Step 3 — Thevenin resistance (short the 20 V source; looking back from the load terminals the 8 Ω, 12 Ω and 4 Ω all appear in parallel):

$$R_{TH}=\frac{1}{\frac{1}{8}+\frac{1}{12}+\frac{1}{4}}=\frac{1}{0.4583}=2.18\ \Omega$$

Step 4 — Load current:

$$I_L=\frac{V_{TH}}{R_{TH}+R_L}=\frac{5.45}{2.18+5}=0.76\ \text{A}$$

Thevenin equivalent circuit: a source $V_{TH}=5.45$ V in series with $R_{TH}=2.18\ \Omega$, feeding the 5 Ω load; the current through the 5 Ω load is 0.76 A.

(a) EMF Equation of a Single-Phase Transformer

Let the flux vary sinusoidally: $\phi=\phi_m\sin\omega t$. The EMF induced per turn is $-\dfrac{d\phi}{dt}$, whose maximum value is $\omega\phi_m=2\pi f\phi_m$.

RMS EMF per turn $=\dfrac{2\pi f\phi_m}{\sqrt2}=4.44\,f\phi_m$.

For $N$ turns:

$$\boxed{E=4.44\,f\,N\,\phi_m}$$

Hence $E_1=4.44 f N_1\phi_m$ and $E_2=4.44 f N_2\phi_m$, giving the turns ratio $\dfrac{E_1}{E_2}=\dfrac{N_1}{N_2}$.

Why a transformer cannot work on DC: A transformer works on mutual induction, which requires a time-varying flux ($e\propto d\phi/dt$). DC produces a constant flux, so $d\phi/dt=0$ and no EMF is induced in the secondary. Moreover, DC is limited only by the small winding resistance, so a heavy current flows and the primary overheats and burns out.

(b) Efficiency Calculation

Given: 25 kVA, 2000/200 V, $R_1=1.5\,\Omega$, $R_2=0.015\,\Omega$, $P_i=350$ W, full load, pf $=0.8$.

Full-load secondary current: $I_2=\dfrac{25000}{200}=125$ A.

Resistance referred to secondary:

$$R_{02}=R_2+R_1\left(\frac{V_2}{V_1}\right)^2=0.015+1.5(0.1)^2=0.030\ \Omega$$

Full-load copper loss:

$$P_{cu}=I_2^2R_{02}=125^2\times0.030=468.75\ \text{W}$$

Output power: $P_o=25000\times0.8=20000$ W.

Efficiency:

$$\eta=\frac{P_o}{P_o+P_{cu}+P_i}=\frac{20000}{20000+468.75+350}=\frac{20000}{20818.75}=96.07\%$$

(a) Series RLC Circuit Relations

Impedance: $Z=R+j(X_L-X_C)$, with $X_L=\omega L$, $X_C=\dfrac{1}{\omega C}$.

Resonance occurs when $X_L=X_C$, i.e. $\omega_0L=\dfrac{1}{\omega_0C}$:

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

At resonance $Z=R$ (minimum) and current is maximum.

Quality factor (voltage magnification):

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

Bandwidth (between the half-power frequencies):

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

(b) Numerical Solution

Given $R=10\,\Omega$, $L=0.1$ H, $C=150\,\mu\text{F}$, $V=200$ V, $f=50$ Hz, $\omega=2\pi(50)=314.16$ rad/s.

$$X_L=\omega L=31.42\ \Omega,\qquad X_C=\frac{1}{\omega C}=21.22\ \Omega$$ $$X=X_L-X_C=10.20\ \Omega\ (\text{inductive})$$

(i) Impedance: $Z=\sqrt{R^2+X^2}=\sqrt{10^2+10.20^2}=14.28\ \Omega$

(ii) Current: $I=\dfrac{V}{Z}=\dfrac{200}{14.28}=14.0\ \text{A}$

(iii) Power factor: $\cos\phi=\dfrac{R}{Z}=\dfrac{10}{14.28}=0.70$ lagging ($\phi=45.6^\circ$, current lags voltage).

(iv) Power consumed: $P=I^2R=14.0^2\times10=1961\ \text{W}\approx1.96\ \text{kW}$.

Phasor diagram (described): Take current $I$ as reference along the horizontal. $V_R=IR$ is in phase with $I$; $V_L=IX_L$ leads $I$ by $90^\circ$ (up); $V_C=IX_C$ lags $I$ by $90^\circ$ (down). Since $V_L>V_C$, the net reactive voltage points up, and the supply voltage $V$ leads $I$ by $\phi=45.6^\circ$.

(a) DC Shunt Motor — Construction, Principle and Back EMF

Construction (sketch described): The motor has a stationary stator carrying field poles; the field winding (many turns of fine wire) is connected in parallel (shunt) with the armature across the supply. The rotating armature carries lap/wave windings in slots, connected to a commutator on which carbon brushes ride. A yoke forms the magnetic frame.

Working principle: When current flows through the armature conductors placed in the field flux, each conductor experiences a force $F=BIl$ (Lorentz/motor action). The commutator reverses the current in each conductor as it passes the brush axis, so the torque on the armature is always unidirectional and the armature rotates.

Back EMF: As the armature rotates in the flux it generates an EMF opposing the supply (Lenz's law), called the back EMF:

$$E_b=\frac{P\phi ZN}{60A}\quad\Rightarrow\quad E_b=V-I_aR_a$$

where $P$=poles, $Z$=conductors, $A$=parallel paths, $\phi$=flux/pole, $N$=speed. Thus $E_b\propto\phi N$.

(b) New Speed when Flux Reduced by 10%

Given $V=220$ V, $I_{a1}=20$ A, $R_a=0.5\,\Omega$, $N_1=1000$ rpm, $\phi_2=0.9\phi_1$, torque constant.

Back EMF (case 1): $E_{b1}=V-I_{a1}R_a=220-20(0.5)=210$ V.

Torque constant: $T\propto\phi I_a$, so $\phi_1 I_{a1}=\phi_2 I_{a2}$:

$$I_{a2}=\frac{\phi_1}{\phi_2}I_{a1}=\frac{I_{a1}}{0.9}=22.22\ \text{A}$$

Back EMF (case 2): $E_{b2}=220-22.22(0.5)=208.89$ V.

New speed (since $N\propto E_b/\phi$):

$$N_2=N_1\cdot\frac{E_{b2}}{E_{b1}}\cdot\frac{\phi_1}{\phi_2}=1000\times\frac{208.89}{210}\times\frac{1}{0.9}=1105\ \text{rpm}$$

The motor speeds up to about 1105 rpm (field weakening raises the speed).

Per-phase impedance: $Z=8+j6\Rightarrow|Z|=\sqrt{8^2+6^2}=10\,\Omega$, $\cos\phi=\dfrac{8}{10}=0.8$ lagging.

Star (Y) Connection

Phase voltage: $V_{ph}=\dfrac{V_L}{\sqrt3}=\dfrac{400}{\sqrt3}=230.94$ V.

Line (= phase) current:

$$I_L=I_{ph}=\frac{V_{ph}}{|Z|}=\frac{230.94}{10}=23.09\ \text{A}$$

Power factor: $\cos\phi=0.8$ lagging.

Total power:

$$P=\sqrt3\,V_L I_L\cos\phi=\sqrt3\times400\times23.09\times0.8=12{,}800\ \text{W}=12.8\ \text{kW}$$

Delta (Δ) Reconnection

In delta each impedance sees the full line voltage, so the phase current and hence the total power are three times the star values:

$$P_\Delta=3\times P_Y=3\times12.8=38.4\ \text{kW}$$

(Check: $I_{ph}=400/10=40$ A, $I_L=\sqrt3\times40=69.3$ A, $P=\sqrt3\times400\times69.3\times0.8=38.4$ kW.)

Given: mean length $l_i=60$ cm $=0.60$ m, air gap $l_g=2$ mm $=0.002$ m, $A=5\,\text{cm}^2=5\times10^{-4}\,\text{m}^2$, $N=500$, $\mu_r=600$, $\phi=0.5\,\text{mWb}=0.5\times10^{-3}$ Wb.

Flux density (same in iron and gap, no fringing):

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

MMF for the iron path (length = $0.60-0.002\approx0.598$ m):

$$AT_i=\frac{B\,l_i}{\mu_0\mu_r}=\frac{1.0\times0.598}{4\pi\times10^{-7}\times600}=793\ \text{AT}$$

MMF for the air gap:

$$AT_g=\frac{B\,l_g}{\mu_0}=\frac{1.0\times0.002}{4\pi\times10^{-7}}=1592\ \text{AT}$$

Total MMF: $AT=793+1592=2385\ \text{AT}$.

Required current:

$$I=\frac{AT}{N}=\frac{2385}{500}=4.77\ \text{A}$$

By the superposition theorem, the current in the 10 Ω resistor is the algebraic sum of the currents due to each source acting alone (the other source replaced by its internal resistance, i.e. short-circuited).

Contribution of the 12 V source (6 V shorted)

The 6 V branch (2 Ω) now appears in parallel with the 10 Ω across the node:

$$10\parallel2=\frac{10\times2}{12}=1.667\,\Omega,\quad V_{node}=12\times\frac{1.667}{4+1.667}=3.53\ \text{V}$$ $$I_{10}'=\frac{3.53}{10}=0.353\ \text{A}$$

Contribution of the 6 V source (12 V shorted)

Now the 12 V branch (4 Ω) is in parallel with the 10 Ω:

$$10\parallel4=\frac{10\times4}{14}=2.857\,\Omega,\quad V_{node}=6\times\frac{2.857}{2+2.857}=3.53\ \text{V}$$ $$I_{10}''=\frac{3.53}{10}=0.353\ \text{A}$$

Total Current

Both sources drive current downward through the 10 Ω in the same direction, so:

$$I_{10}=I_{10}'+I_{10}''=0.353+0.353=0.706\ \text{A}$$

The current through the 10 Ω resistor is ≈ 0.706 A.

(a) PMMC Instrument

Principle: A light coil pivoted in the field of a permanent magnet carries the current to be measured. The current-carrying coil in the radial magnetic field experiences a deflecting torque $T_d=BINA\propto I$. Spiral hairsprings provide the controlling torque ($T_c\propto\theta$); at balance $\theta\propto I$, giving a uniform (linear) scale. Eddy currents in the metal former give damping.

Why unsuitable for AC: The deflecting torque reverses with the current. On AC the torque alternates so rapidly that the pointer (due to inertia) reads the average torque, which is zero for a symmetrical AC wave. Hence a PMMC reads only DC (or the DC component); it needs a rectifier to measure AC.

(b) Shunt Resistance for Range Extension

Given $I_m=1$ mA (full-scale), $R_m=100\,\Omega$, required range $I=5$ A.

Shunt current: $I_{sh}=I-I_m=5-0.001=4.999$ A.

Voltage across coil = voltage across shunt:

$$R_{sh}=\frac{I_m R_m}{I-I_m}=\frac{0.001\times100}{4.999}=0.0200\ \Omega$$

The required shunt resistance is ≈ 0.02 Ω (multiplying factor $n=5000$, so $R_{sh}=R_m/(n-1)=100/4999=0.02\,\Omega$).

(a) Rotating Magnetic Field

When a balanced three-phase supply feeds the three stator windings (displaced $120^\circ$ in space and carrying currents $120^\circ$ apart in time), the resultant of the three pulsating fluxes is a flux of constant magnitude ($1.5\,\phi_m$) that rotates around the air gap at synchronous speed $N_s=\dfrac{120f}{P}$. This rotating field sweeps past the rotor conductors, induces EMF and current in them (Faraday's law), and the interaction of rotor current with the field produces torque that drags the rotor in the direction of field rotation (Lenz's law / motor action).

(b) Numerical Solution

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

Synchronous speed:

$$N_s=\frac{120f}{P}=\frac{120\times50}{4}=1500\ \text{rpm}$$

Slip:

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

Rotor (slip) frequency:

$$f_r=s\,f=0.04\times50=2\ \text{Hz}$$

Definitions

RMS value: the equivalent steady DC value that produces the same heating (power) in a given resistance as the alternating quantity; $I_{rms}=\sqrt{\dfrac{1}{T}\int_0^T i^2\,dt}$.

Average value: the arithmetic mean of the instantaneous values over a half cycle (for a symmetrical wave the full-cycle average is zero); $I_{avg}=\dfrac{1}{T/2}\int_0^{T/2} i\,dt$.

For $i(t)=50\sin(314t)$ A

Peak $I_m=50$ A, $\omega=314$ rad/s.

RMS value: $I_{rms}=\dfrac{I_m}{\sqrt2}=\dfrac{50}{\sqrt2}=35.36\ \text{A}$

Average value (half cycle): $I_{avg}=\dfrac{2I_m}{\pi}=\dfrac{2\times50}{\pi}=31.83\ \text{A}$

Form factor: $\dfrac{I_{rms}}{I_{avg}}=\dfrac{35.36}{31.83}=1.11$

Peak (crest) factor: $\dfrac{I_m}{I_{rms}}=\dfrac{50}{35.36}=1.414$

Frequency: $f=\dfrac{\omega}{2\pi}=\dfrac{314}{2\pi}=49.97\approx50\ \text{Hz}$

Open-Circuit (No-Load) Test

Performed usually on the LV side with the HV side open. Rated voltage at rated frequency is applied; the secondary is open so only the small no-load (magnetising) current flows.

• Since the no-load current is small, copper losses are negligible; the wattmeter reading $W_0$ gives the iron (core) loss (constant at all loads).
• From $W_0,V_0,I_0$: no-load pf $\cos\phi_0=\dfrac{W_0}{V_0 I_0}$; magnetising branch parameters $R_0=\dfrac{V_0}{I_0\cos\phi_0}$ (core-loss resistance) and $X_0=\dfrac{V_0}{I_0\sin\phi_0}$ (magnetising reactance).

Short-Circuit (Impedance) Test

Performed on the HV side with the LV side short-circuited. A reduced voltage ($V_{sc}$, a few % of rated) is applied so that rated current flows.

• The flux (and hence iron loss) is very small; the wattmeter reading $W_{sc}$ gives the full-load copper loss.
• Equivalent parameters referred to the supply side: $Z_{eq}=\dfrac{V_{sc}}{I_{sc}}$, $R_{eq}=\dfrac{W_{sc}}{I_{sc}^2}$, $X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}$.

Summary

• OC test → iron loss and shunt branch ($R_0,X_0$).
• SC test → copper loss and series branch ($R_{eq},X_{eq}$).

Together they give the complete approximate equivalent circuit and allow efficiency and voltage regulation to be predicted without loading the transformer.

Generated EMF

Given: $P_o=50$ kW, $V=250$ V, $R_a=0.02\,\Omega$, $R_{sh}=50\,\Omega$.

Load current: $I_L=\dfrac{P_o}{V}=\dfrac{50000}{250}=200$ A.

Shunt field current: $I_{sh}=\dfrac{V}{R_{sh}}=\dfrac{250}{50}=5$ A.

Armature current: $I_a=I_L+I_{sh}=200+5=205$ A.

Generated EMF:

$$E_g=V+I_aR_a=250+205\times0.02=250+4.1=254.1\ \text{V}$$

External Characteristic of a DC Shunt Generator

The external characteristic is the graph of terminal voltage $V$ versus load current $I_L$ at constant speed.

Shape (described): Starting from the no-load voltage, $V$ droops slightly as load increases, because (i) the armature drop $I_aR_a$ rises, (ii) armature reaction weakens the flux, and (iii) the reduced terminal voltage lowers the field current and hence the EMF. The drop is gradual and the generator is fairly stable up to full load. Beyond a critical (breakdown) load current the curve turns back (current decreases as load resistance is reduced further), so the generator has a self-protecting tendency against short circuits.