BE Civil Engineering (IOE, TU) Basic Electrical Engineering (IOE, EE 401) Question Paper 2078 Nepal

Setup. Take ground as reference (0 V). Let $V_A$ and $V_B$ be the node voltages.

(a) Nodal equations.

At node A (KCL, currents leaving):

$$\frac{V_A-24}{4}+\frac{V_A}{6}+\frac{V_A-V_B}{3}=0$$

Multiply by 12:

$$3(V_A-24)+2V_A+4(V_A-V_B)=0$$ $$3V_A-72+2V_A+4V_A-4V_B=0 \Rightarrow 9V_A-4V_B=72 \quad (1)$$

At node B:

$$\frac{V_B-V_A}{3}+\frac{V_B}{8}+\frac{V_B-12}{2}=0$$

Multiply by 24:

$$8(V_B-V_A)+3V_B+12(V_B-12)=0$$ $$8V_B-8V_A+3V_B+12V_B-144=0 \Rightarrow -8V_A+23V_B=144 \quad (2)$$

Solve. From (1): $V_A=\dfrac{72+4V_B}{9}$. Substitute into (2):

$$-8\left(\frac{72+4V_B}{9}\right)+23V_B=144$$

Multiply by 9:

$$-8(72+4V_B)+207V_B=1296$$ $$-576-32V_B+207V_B=1296 \Rightarrow 175V_B=1872 \Rightarrow V_B=10.697\ \text{V}$$

Then $V_A=\dfrac{72+4(10.697)}{9}=\dfrac{72+42.789}{9}=12.754\ \text{V}$.

$$\boxed{V_A \approx 12.75\ \text{V}, \quad V_B \approx 10.70\ \text{V}}$$

(b) Current through the $3\ \Omega$ resistor.

$$I_{3}=\frac{V_A-V_B}{3}=\frac{12.754-10.697}{3}=\frac{2.057}{3}=0.686\ \text{A}$$

Since $V_A>V_B$, current flows from A to B.

$$\boxed{I_{3}\approx 0.686\ \text{A, from A to B}}$$

(c) Power delivered by the $24\ \text{V}$ source. Current from the source through the $4\ \Omega$ resistor into node A:

$$I_{src}=\frac{24-V_A}{4}=\frac{24-12.754}{4}=\frac{11.246}{4}=2.811\ \text{A}$$

Power delivered by the 24 V source:

$$P=24\times I_{src}=24\times 2.811=67.5\ \text{W}$$ $$\boxed{P_{24V}\approx 67.5\ \text{W}}$$

(a) Thevenin voltage $V_{Th}$ (open-circuit voltage at a-b).

With a-b open, no current leaves the terminals. Apply nodal analysis at node a (b = 0):

$$\frac{V_a-40}{5}+\frac{V_a}{10}-4=0$$

(The $4\ \text{A}$ source injects current into node a, hence the $-4$.) Multiply by 10:

$$2(V_a-40)+V_a-40=0 \Rightarrow 2V_a-80+V_a-40=0 \Rightarrow 3V_a=120$$ $$V_a=40\ \text{V} \Rightarrow V_{Th}=40\ \text{V}$$

(b) Thevenin resistance $R_{Th}$. Deactivate sources: short the $40\ \text{V}$ source, open the $4\ \text{A}$ source. Then from a-b we see the $5\ \Omega$ (now to ground) in parallel with the $10\ \Omega$:

$$R_{Th}=\frac{5\times 10}{5+10}=\frac{50}{15}=3.33\ \Omega$$

For maximum power transfer:

$$\boxed{R_L=R_{Th}=3.33\ \Omega}$$

(c) Maximum power.

$$P_{max}=\frac{V_{Th}^2}{4R_{Th}}=\frac{40^2}{4\times 3.33}=\frac{1600}{13.33}=120\ \text{W}$$ $$\boxed{P_{max}=120\ \text{W}}$$

Note: $V_{Th}=40\ \text{V}$, $R_{Th}=3.33\ \Omega$, $R_L=3.33\ \Omega$, $P_{max}=120\ \text{W}$.

(a) Reactances and impedance.

Angular frequency: $\omega = 2\pi f = 2\pi(50)=314.16\ \text{rad/s}$.

Inductive reactance:

$$X_L=\omega L=314.16\times 0.08=25.13\ \Omega$$

Capacitive reactance:

$$X_C=\frac{1}{\omega C}=\frac{1}{314.16\times 120\times10^{-6}}=\frac{1}{0.0377}=26.53\ \Omega$$

Net reactance:

$$X=X_L-X_C=25.13-26.53=-1.40\ \Omega \ (\text{capacitive})$$

Impedance:

$$Z=\sqrt{R^2+X^2}=\sqrt{12^2+(-1.40)^2}=\sqrt{144+1.96}=\sqrt{145.96}=12.08\ \Omega$$ $$\boxed{X_L=25.13\ \Omega,\ X_C=26.53\ \Omega,\ Z=12.08\ \Omega}$$

(b) Current and power factor.

$$I=\frac{V}{Z}=\frac{230}{12.08}=19.04\ \text{A}$$

Power factor:

$$\cos\phi=\frac{R}{Z}=\frac{12}{12.08}=0.9932$$

Since $X_C>X_L$ (net capacitive), the current leads the voltage.

$$\phi=\cos^{-1}(0.9932)=6.66^\circ \ \text{leading}$$ $$\boxed{I=19.04\ \text{A},\ \text{pf}=0.993\ \text{leading}}$$

(c) Powers. Real power:

$$P=VI\cos\phi=230\times 19.04\times 0.9932=4349\ \text{W}=4.35\ \text{kW}$$

Reactive power:

$$Q=VI\sin\phi=230\times 19.04\times \sin(6.66^\circ)=230\times 19.04\times 0.1160=508\ \text{VAR (capacitive)}$$

Apparent power:

$$S=VI=230\times 19.04=4379\ \text{VA}=4.38\ \text{kVA}$$ $$\boxed{P\approx 4.35\ \text{kW},\ Q\approx 0.51\ \text{kVAR (cap.)},\ S\approx 4.38\ \text{kVA}}$$

(a) Star connection.

Phase voltage:

$$V_{ph}=\frac{V_L}{\sqrt3}=\frac{400}{1.732}=230.9\ \text{V}$$

Impedance magnitude:

$$|Z|=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\ \Omega$$

Phase current (= line current in star):

$$I_{ph}=\frac{V_{ph}}{|Z|}=\frac{230.9}{10}=23.09\ \text{A}=I_L$$ $$\boxed{V_{ph}=230.9\ \text{V},\ I_L=23.09\ \text{A}}$$

(b) Power factor and total power.

$$\cos\phi=\frac{R}{|Z|}=\frac{8}{10}=0.8\ \text{lagging}$$

Total real power:

$$P_Y=\sqrt3\,V_L I_L\cos\phi=1.732\times 400\times 23.09\times 0.8=12{,}800\ \text{W}=12.8\ \text{kW}$$

(Check: $P=3I_{ph}^2R=3\times 23.09^2\times 8=3\times 533.1\times 8=12{,}794\ \text{W}$ ✓)

$$\boxed{\text{pf}=0.8\ \text{lagging},\ P_Y=12.8\ \text{kW}}$$

(c) Delta connection (same supply).

In delta, phase voltage = line voltage = $400\ \text{V}$.

$$I_{ph,\Delta}=\frac{V_L}{|Z|}=\frac{400}{10}=40\ \text{A}$$

Line current:

$$I_{L,\Delta}=\sqrt3\,I_{ph,\Delta}=1.732\times 40=69.28\ \text{A}$$

Total real power:

$$P_\Delta=\sqrt3\,V_L I_{L,\Delta}\cos\phi=1.732\times 400\times 69.28\times 0.8=38{,}400\ \text{W}=38.4\ \text{kW}$$

Comment. $\dfrac{P_\Delta}{P_Y}=\dfrac{38.4}{12.8}=3$. A delta-connected load draws three times the power (and three times the line current) of the same impedances in star across the same supply.

$$\boxed{I_{L,\Delta}=69.28\ \text{A},\ P_\Delta=38.4\ \text{kW},\ P_\Delta/P_Y=3}$$

(a) Losses.

Iron loss = open-circuit power input (since copper loss is negligible at no load):

$$P_i=105\ \text{W}$$

The short-circuit test was done at HV rated current. Full-load HV current:

$$I_{HV,FL}=\frac{25{,}000}{2000}=12.5\ \text{A}$$

The SC test current is exactly $12.5\ \text{A}$ = full load, so the measured power is the full-load copper loss:

$$P_{cu,FL}=360\ \text{W}$$ $$\boxed{P_i=105\ \text{W},\quad P_{cu,FL}=360\ \text{W}}$$

(b) Efficiency.

Full-load output power at 0.8 pf:

$$P_{out,FL}=25{,}000\times 0.8=20{,}000\ \text{W}$$

Total losses at full load $=P_i+P_{cu,FL}=105+360=465\ \text{W}$.

$$\eta_{FL}=\frac{20{,}000}{20{,}000+465}=\frac{20{,}000}{20{,}465}=0.9773=97.73\%$$

At half load ($x=0.5$): copper loss scales as $x^2$:

$$P_{cu}=0.5^2\times 360=0.25\times 360=90\ \text{W}$$

Output $=0.5\times 20{,}000=10{,}000\ \text{W}$; losses $=105+90=195\ \text{W}$.

$$\eta_{HL}=\frac{10{,}000}{10{,}000+195}=\frac{10{,}000}{10{,}195}=0.9809=98.09\%$$ $$\boxed{\eta_{FL}=97.73\%,\quad \eta_{HL}=98.09\%}$$

(c) Load for maximum efficiency.

Maximum efficiency occurs when copper loss = iron loss, i.e. $x^2 P_{cu,FL}=P_i$:

$$x=\sqrt{\frac{P_i}{P_{cu,FL}}}=\sqrt{\frac{105}{360}}=\sqrt{0.2917}=0.540$$

So maximum efficiency occurs at about 54% of full load, i.e. a load of $0.540\times 25=13.5\ \text{kVA}$.

$$\boxed{x_{max}=0.540\ (\approx 54\%\ \text{of full load},\ 13.5\ \text{kVA})}$$

(a) Equivalent capacitance (series).

$$\frac{1}{C_{eq}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{3+2+1}{12}=\frac{6}{12}=\frac{1}{2}$$ $$C_{eq}=2\ \mu\text{F}$$

(b) Charge and voltages. In series, charge is the same on each capacitor:

$$Q=C_{eq}V=2\times10^{-6}\times 200=4\times10^{-4}\ \text{C}=400\ \mu\text{C}$$

Voltage across each:

$$V_4=\frac{Q}{C_4}=\frac{400}{4}=100\ \text{V}$$ $$V_6=\frac{400}{6}=66.67\ \text{V}$$ $$V_{12}=\frac{400}{12}=33.33\ \text{V}$$

Check: $100+66.67+33.33=200\ \text{V}$ ✓

$$\boxed{Q=400\ \mu\text{C};\ V_4=100\ \text{V},\ V_6=66.67\ \text{V},\ V_{12}=33.33\ \text{V}}$$

(c) Total energy stored.

$$E=\frac12 C_{eq}V^2=\frac12\times 2\times10^{-6}\times 200^2=\frac12\times 2\times10^{-6}\times 40{,}000=0.04\ \text{J}$$ $$\boxed{E=0.04\ \text{J}=40\ \text{mJ}}$$

(a) Reluctances.

Cross-sectional area: $A=5\ \text{cm}^2=5\times10^{-4}\ \text{m}^2$.

Iron path length (ring minus gap): $l_{iron}=0.6-0.001=0.599\ \text{m}$.

Reluctance of iron:

$$S_{iron}=\frac{l_{iron}}{\mu_0\mu_r A}=\frac{0.599}{(4\pi\times10^{-7})(1200)(5\times10^{-4})}$$

Denominator $=4\pi\times10^{-7}\times1200\times5\times10^{-4}=7.540\times10^{-7}$.

$$S_{iron}=\frac{0.599}{7.540\times10^{-7}}=7.944\times10^{5}\ \text{AT/Wb}$$

Reluctance of air gap:

$$S_{gap}=\frac{l_g}{\mu_0 A}=\frac{0.001}{(4\pi\times10^{-7})(5\times10^{-4})}=\frac{0.001}{6.283\times10^{-10}}=1.592\times10^{6}\ \text{AT/Wb}$$ $$\boxed{S_{iron}=7.94\times10^{5}\ \text{AT/Wb},\ S_{gap}=1.59\times10^{6}\ \text{AT/Wb}}$$

(b) Required MMF. Total reluctance:

$$S=S_{iron}+S_{gap}=7.944\times10^5+1.592\times10^6=2.386\times10^6\ \text{AT/Wb}$$

MMF for flux $\Phi=0.8\times10^{-3}\ \text{Wb}$:

$$\mathcal{F}=\Phi\,S=0.8\times10^{-3}\times 2.386\times10^6=1909\ \text{AT}$$ $$\boxed{\mathcal{F}\approx 1909\ \text{ampere-turns}}$$

(c) Coil current.

$$I=\frac{\mathcal{F}}{N}=\frac{1909}{400}=4.77\ \text{A}$$ $$\boxed{I\approx 4.77\ \text{A}}$$

(a) Basic parameters. Comparing with $i=I_m\sin(\omega t)$:

• Peak value: $I_m=15\ \text{A}$
• Angular frequency: $\omega=314\ \text{rad/s}$
• Frequency: $f=\dfrac{\omega}{2\pi}=\dfrac{314}{6.283}=49.97\approx 50\ \text{Hz}$

(b) RMS and average values.

$$I_{rms}=\frac{I_m}{\sqrt2}=\frac{15}{1.414}=10.61\ \text{A}$$ $$I_{avg}=\frac{2I_m}{\pi}=\frac{2\times15}{3.1416}=\frac{30}{3.1416}=9.55\ \text{A}$$ $$\boxed{I_{rms}=10.61\ \text{A},\ I_{avg}=9.55\ \text{A}}$$

(c) Form factor and peak factor.

$$\text{Form factor}=\frac{I_{rms}}{I_{avg}}=\frac{10.61}{9.55}=1.11$$ $$\text{Peak factor}=\frac{I_m}{I_{rms}}=\frac{15}{10.61}=1.414$$ $$\boxed{\text{FF}=1.11,\ \text{PF}=1.414}$$

(d) Instantaneous current at $t=4\ \text{ms}$.

$$\theta=\omega t=314\times 0.004=1.256\ \text{rad}\ (=71.96^\circ)$$ $$i=15\sin(1.256)=15\times 0.9510=14.27\ \text{A}$$ $$\boxed{i(4\ \text{ms})\approx 14.27\ \text{A}}$$

(a) Field and armature currents. Field current:

$$I_{sh}=\frac{V}{R_{sh}}=\frac{220}{110}=2\ \text{A}$$

Armature current:

$$I_a=I_L-I_{sh}=42-2=40\ \text{A}$$ $$\boxed{I_{sh}=2\ \text{A},\ I_a=40\ \text{A}}$$

(b) Back EMF at full load.

$$E_b=V-I_aR_a=220-(40)(0.25)=220-10=210\ \text{V}$$ $$\boxed{E_b=210\ \text{V}}$$

(c) Full-load speed. At no load, armature current $=4\ \text{A}$:

$$E_{b0}=V-I_{a0}R_a=220-(4)(0.25)=220-1=219\ \text{V}$$

With constant flux, $N\propto E_b$:

$$\frac{N_{FL}}{N_0}=\frac{E_b}{E_{b0}}=\frac{210}{219}$$ $$N_{FL}=1000\times\frac{210}{219}=1000\times 0.9589=958.9\ \text{rpm}$$ $$\boxed{N_{FL}\approx 959\ \text{rpm}}$$

Kirchhoff's Current Law (KCL).

Statement: The algebraic sum of currents at any node (junction) of an electric circuit is zero; equivalently, the total current entering a node equals the total current leaving it.

$$\sum_{k} I_k = 0 \quad\text{(currents in } = \text{currents out)}$$

Physical basis: Conservation of electric charge — charge cannot accumulate at a node, so the rate of charge in must equal the rate of charge out.

Illustration: At a node, currents $I_1=5\ \text{A}$ and $I_2=3\ \text{A}$ flow in, and $I_3$ flows out. Then:

$$I_1+I_2-I_3=0 \Rightarrow I_3=5+3=8\ \text{A}$$

Kirchhoff's Voltage Law (KVL).

Statement: The algebraic sum of all voltages (EMF rises and resistive drops) around any closed loop of a circuit is zero.

$$\sum_{k} V_k = 0 \quad\Rightarrow\quad \sum \text{EMFs} = \sum \text{IR drops}$$

Physical basis: Conservation of energy — a unit charge taken around a closed loop returns to its starting potential, so net energy change per unit charge is zero.

Together, KCL and KVL form the basis of nodal and mesh analysis of circuits.

(b) Superposition Theorem.

Statement: In any linear network containing two or more independent sources, 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.

Steps to apply:

1. Select one independent source; deactivate all others — replace voltage sources by short circuits and current sources by open circuits (leave internal resistances in place).
2. Compute the desired current/voltage due to the active source alone.
3. Repeat for every independent source in turn.
4. Algebraically add all the individual responses to obtain the total response.

Limitation: It applies only to linear circuits and only to current/voltage (linear quantities); it cannot be used directly to compute power, since power is a nonlinear ($I^2R$) function of current/voltage.

Working principle. A transformer works on the principle of mutual electromagnetic induction (Faraday's law). An alternating voltage applied to the primary winding sets up an alternating flux in the laminated iron core. This time-varying flux links the secondary winding and induces an EMF in it. There is no electrical connection between the windings — energy is transferred magnetically. Since both windings link (ideally) the same flux, the induced EMF per turn is equal, giving $\dfrac{E_1}{E_2}=\dfrac{N_1}{N_2}$.

EMF equation derivation. Let the flux be $\phi=\phi_m\sin(\omega t)$, with $\omega=2\pi f$. EMF induced in a winding of $N$ turns (Faraday's law):

$$e=-N\frac{d\phi}{dt}=-N\phi_m\omega\cos(\omega t)$$

Maximum EMF: $E_m=N\phi_m\omega=2\pi f N\phi_m$. RMS value:

$$E=\frac{E_m}{\sqrt2}=\frac{2\pi f N\phi_m}{\sqrt2}=4.44\,f\,N\,\phi_m$$ $$\boxed{E=4.44\,f\,N\,\phi_m}$$

Numerical calculation.

$$E_1=4.44\times f\times N_1\times \phi_m=4.44\times 50\times 500\times 4\times10^{-3}$$ $$E_1=4.44\times 50\times 500\times 0.004=4.44\times 100=444\ \text{V}$$ $$\boxed{E_1=444\ \text{V}}$$