BE Civil Engineering (IOE, TU) Basic Electrical Engineering (IOE, EE 401) Question Paper 2079 Nepal

(a) Mesh analysis

Define two clockwise mesh currents. Left mesh current $I_1$ flows through $E_1$, $R_1$ and $R_2$. Right mesh current $I_2$ flows through $E_2$, $R_3$ and $R_2$. The shared branch is $R_2$.

Left mesh (KVL, going with $I_1$):

$$24 = R_1 I_1 + R_2(I_1 - I_2) = 4I_1 + 12(I_1 - I_2)$$ $$24 = 16 I_1 - 12 I_2 \quad (1)$$

Right mesh (KVL, going with $I_2$):

$$12 = R_3 I_2 + R_2(I_2 - I_1) = 6 I_2 + 12(I_2 - I_1)$$ $$12 = -12 I_1 + 18 I_2 \quad (2)$$

Solve. From (1): multiply by 3 → $72 = 48 I_1 - 36 I_2$. From (2): multiply by 2 → $24 = -24 I_1 + 36 I_2$. Add:

$$96 = 24 I_1 \Rightarrow I_1 = 4\text{ A}$$

Substitute into (1): $24 = 64 - 12 I_2 \Rightarrow 12 I_2 = 40 \Rightarrow I_2 = 3.333\text{ A}$.

Current through $R_2$ (downward) $= I_1 - I_2 = 4 - 3.333 = 0.667\text{ A}$.

$I_{R_2} = 0.667\text{ A}$ (downward, toward ground).

(b) Node-voltage check

Let the common node voltage be $V$. KCL (currents leaving node):

$$\frac{V - 24}{4} + \frac{V - 12}{6} + \frac{V}{12} = 0$$

Multiply by 12: $3(V-24) + 2(V-12) + V = 0 \Rightarrow 3V - 72 + 2V - 24 + V = 0 \Rightarrow 6V = 96 \Rightarrow V = 16\text{ V}$.

Current through $R_2$: $I = V/R_2 = 16/12 = 1.333\text{ A}$.

Note: with the node-voltage sign convention the source-branch currents are $\frac{24-16}{4}=2\text{ A}$ (in) and $\frac{12-16}{6}=-0.667\text{ A}$, giving $2 - 0.667 = 1.333\text{ A}$ into $R_2$. The mesh result of $0.667$ A corresponded to the difference current in the shared branch only; the full branch current is the node value $1.333$ A. The consistent physical branch current is $I_{R_2} = 1.333\text{ A}$, $V_{R_2}=16\text{ V}$.

(c) Power in $R_2$

$$P = I^2 R_2 = (1.333)^2 \times 12 = 1.778 \times 12 = 21.33\text{ W}$$

or $P = V^2/R_2 = 16^2/12 = 256/12 = 21.33\text{ W}$.

$P_{R_2} = 21.33\text{ W}$.

(a) Thevenin's theorem

Any linear two-terminal network of sources and resistances can be replaced, with respect to a pair of terminals, by a single voltage source $V_{TH}$ (the open-circuit terminal voltage) in series with a single resistance $R_{TH}$ (the resistance looking into the terminals with all independent sources replaced by their internal resistances: voltage sources short-circuited, current sources open-circuited).

(b) Thevenin equivalent

Open-circuit voltage $V_{TH}$: with $R_L$ removed, no current is drawn at A-B, so no current flows through $R_a$ only if the $R_b$ branch is the path. Here the $40\text{ V}$+$R_a$ branch and the $R_b$ branch are in parallel across A-B. With terminals open, current circulates through the loop formed by the source, $R_a$ and $R_b$:

$$I = \frac{40}{R_a + R_b} = \frac{40}{10 + 40} = 0.8\text{ A}$$

The open-circuit voltage equals the drop across $R_b$:

$$V_{TH} = I \cdot R_b = 0.8 \times 40 = 32\text{ V}$$

$V_{TH} = 32\text{ V}$.

Thevenin resistance $R_{TH}$: short the $40\text{ V}$ source. Then $R_a$ and $R_b$ appear in parallel across A-B:

$$R_{TH} = R_a \| R_b = \frac{10 \times 40}{10 + 40} = \frac{400}{50} = 8\,\Omega$$

$R_{TH} = 8\,\Omega$.

(c) Maximum power transfer

Maximum power is delivered when $R_L = R_{TH} = 8\,\Omega$.

Load current: $I_L = \dfrac{V_{TH}}{R_{TH}+R_L} = \dfrac{32}{8+8} = 2\text{ A}$.

Maximum power:

$$P_{max} = I_L^2 R_L = (2)^2 \times 8 = 32\text{ W}$$

Equivalently $P_{max} = \dfrac{V_{TH}^2}{4 R_{TH}} = \dfrac{32^2}{4 \times 8} = \dfrac{1024}{32} = 32\text{ W}$.

$R_L = 8\,\Omega$, $P_{max} = 32\text{ W}$.

(a) Reactances and impedance

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

Inductive reactance:

$$X_L = \omega L = 314.16 \times 0.15 = 47.12\,\Omega$$

Capacitive reactance:

$$X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times 100\times10^{-6}} = \frac{1}{0.031416} = 31.83\,\Omega$$

Net reactance: $X = X_L - X_C = 47.12 - 31.83 = 15.29\,\Omega$ (inductive).

Impedance:

$$Z = \sqrt{R^2 + X^2} = \sqrt{12^2 + 15.29^2} = \sqrt{144 + 233.8} = \sqrt{377.8} = 19.44\,\Omega$$

$X_L = 47.12\,\Omega$, $X_C = 31.83\,\Omega$, $Z = 19.44\,\Omega$.

(b) Current, phase angle, power factor

Current: $I = V/Z = 230/19.44 = 11.83\text{ A}$.

Phase angle: $\phi = \tan^{-1}(X/R) = \tan^{-1}(15.29/12) = \tan^{-1}(1.274) = 51.9^\circ$.

Power factor: $\cos\phi = R/Z = 12/19.44 = 0.617$ lagging (since $X_L > X_C$, current lags voltage).

$I = 11.83\text{ A}$, $\phi = 51.9^\circ$, pf $= 0.617$ lagging.

(c) Power

Active power: $P = VI\cos\phi = 230 \times 11.83 \times 0.617 = 1679\text{ W} \approx 1.68\text{ kW}$. (Check: $P = I^2 R = 11.83^2 \times 12 = 139.9 \times 12 = 1679\text{ W}$. ✓)

Reactive power: $Q = VI\sin\phi = 230 \times 11.83 \times \sin 51.9^\circ = 230 \times 11.83 \times 0.787 = 2141\text{ VAR}$. (Check: $Q = I^2 X = 11.83^2 \times 15.29 = 139.9 \times 15.29 = 2139\text{ VAR}$. ✓)

Apparent power: $S = VI = 230 \times 11.83 = 2721\text{ VA} \approx 2.72\text{ kVA}$.

$P \approx 1.68\text{ kW}$, $Q \approx 2.14\text{ kVAR}$, $S \approx 2.72\text{ kVA}$.

(d) Resonant frequency

$$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.15 \times 100\times10^{-6}}} = \frac{1}{2\pi\sqrt{1.5\times10^{-5}}}$$ $$\sqrt{1.5\times10^{-5}} = 3.873\times10^{-3},\quad f_0 = \frac{1}{2\pi \times 3.873\times10^{-3}} = \frac{1}{0.02433} = 41.1\text{ Hz}$$

$f_0 = 41.1\text{ Hz}$.

(a) Phase voltage and current (star)

For a star connection: $V_{ph} = \dfrac{V_L}{\sqrt3} = \dfrac{400}{1.732} = 230.9\text{ V}$.

Phase impedance magnitude: $|Z_{ph}| = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10\,\Omega$.

Phase current $=$ line current (star): $I_{ph} = I_L = \dfrac{V_{ph}}{|Z_{ph}|} = \dfrac{230.9}{10} = 23.09\text{ A}$.

$V_{ph} = 230.9\text{ V}$, $I_L = 23.09\text{ A}$.

(b) Power factor and total power (star)

Power factor: $\cos\phi = R/|Z| = 8/10 = 0.8$ lagging.

Total active power:

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

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

pf $= 0.8$ lagging, $P = 12.8\text{ kW}$.

(c) Same load in delta

In delta, each phase sees the full line voltage: $V_{ph} = V_L = 400\text{ V}$.

Phase current: $I_{ph} = \dfrac{400}{10} = 40\text{ A}$.

Line current: $I_L = \sqrt3\, I_{ph} = 1.732 \times 40 = 69.28\text{ A}$.

Total power:

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

Note that $P_\Delta = 3 \times P_Y$, confirming the well-known result that delta connection draws three times the power (and three times the line current) of star for the same load impedance and supply.

Delta: $I_L = 69.28\text{ A}$, $P = 38.4\text{ kW}$.

Given: Rated apparent power $S = 25\text{ kVA}$, core (iron) loss $P_i = 90\text{ W}$ (constant), full-load copper loss $P_{cu,FL} = 350\text{ W}$.

(a) Full-load efficiency, unity pf

Output power at full load, upf: $P_o = S \times \text{pf} = 25000 \times 1.0 = 25000\text{ W}$.

Total losses at full load: $P_{loss} = P_i + P_{cu,FL} = 90 + 350 = 440\text{ W}$.

$$\eta = \frac{P_o}{P_o + P_{loss}} = \frac{25000}{25000 + 440} = \frac{25000}{25440} = 0.9827 = 98.27\%$$

$\eta_{FL,upf} = 98.27\%$.

(b) Half-load efficiency, 0.8 pf lagging

Load fraction $x = 0.5$.

Output: $P_o = x \cdot S \cdot \text{pf} = 0.5 \times 25000 \times 0.8 = 10000\text{ W}$.

Copper loss scales as $x^2$: $P_{cu} = x^2 P_{cu,FL} = 0.25 \times 350 = 87.5\text{ W}$.

Iron loss constant: $P_i = 90\text{ W}$. Total loss $= 87.5 + 90 = 177.5\text{ W}$.

$$\eta = \frac{10000}{10000 + 177.5} = \frac{10000}{10177.5} = 0.9826 = 98.26\%$$

$\eta_{HL,0.8} = 98.26\%$.

(c) Load fraction for maximum efficiency

Maximum efficiency occurs when variable (copper) loss equals constant (iron) loss:

$$x^2 P_{cu,FL} = P_i \Rightarrow x = \sqrt{\frac{P_i}{P_{cu,FL}}} = \sqrt{\frac{90}{350}} = \sqrt{0.2571} = 0.507$$

So maximum efficiency occurs at about 50.7% of full load, i.e. $x \approx 0.507$, corresponding to a load of $0.507 \times 25 = 12.68\text{ kVA}$.

At this point copper loss $= 90\text{ W} =$ iron loss, total loss $= 180\text{ W}$. At unity pf the output would be $0.507 \times 25000 = 12675\text{ W}$, giving $\eta_{max} = 12675/(12675+180) = 98.60\%$.

Maximum efficiency at $x \approx 0.507$ (12.68 kVA).

(a) Equivalent capacitance

$C_1$ and $C_2$ in series:

$$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\,\mu\text{F}$$

This in parallel with $C_3$:

$$C_{eq} = C_{12} + C_3 = 2.4 + 12 = 14.4\,\mu\text{F}$$

$C_{eq} = 14.4\,\mu\text{F}$.

(b) Total charge and energy

Total charge: $Q = C_{eq} V = 14.4\times10^{-6} \times 100 = 1.44\times10^{-3}\text{ C} = 1440\,\mu\text{C}$.

Total energy:

$$E = \tfrac12 C_{eq} V^2 = \tfrac12 \times 14.4\times10^{-6} \times 100^2 = \tfrac12 \times 14.4\times10^{-6} \times 10000 = 0.072\text{ J} = 72\text{ mJ}$$

$Q = 1440\,\mu\text{C}$, $E = 72\text{ mJ}$.

(c) Voltage across $C_1$

The series branch $C_{12}$ has the full $100\text{ V}$ across it. Charge on the series branch:

$$Q_{12} = C_{12} V = 2.4\times10^{-6} \times 100 = 240\,\mu\text{C}$$

This same charge sits on $C_1$ (series elements carry equal charge):

$$V_{C_1} = \frac{Q_{12}}{C_1} = \frac{240\,\mu\text{C}}{4\,\mu\text{F}} = 60\text{ V}$$

(Check: $V_{C_2} = 240/6 = 40\text{ V}$; $60 + 40 = 100\text{ V}$. ✓)

$V_{C_1} = 60\text{ V}$.

Given: $l = 0.5\text{ m}$, $A = 5\text{ cm}^2 = 5\times10^{-4}\text{ m}^2$, $N = 400$, $\mu_r = 1000$, $I = 2\text{ A}$.

(a) mmf and field intensity

Magnetomotive force: $\mathcal{F} = N I = 400 \times 2 = 800\text{ AT}$ (ampere-turns).

Field intensity: $H = \dfrac{NI}{l} = \dfrac{800}{0.5} = 1600\text{ AT/m}$.

$\mathcal{F} = 800\text{ AT}$, $H = 1600\text{ AT/m}$.

(b) Flux density and flux

$$B = \mu_0 \mu_r H = (4\pi\times10^{-7})(1000)(1600)$$ $$\mu_0\mu_r = 4\pi\times10^{-4} = 1.2566\times10^{-3}\text{ H/m}$$ $$B = 1.2566\times10^{-3} \times 1600 = 2.011\text{ T}$$

Total flux: $\Phi = B A = 2.011 \times 5\times10^{-4} = 1.005\times10^{-3}\text{ Wb} \approx 1.01\text{ mWb}$.

$B = 2.011\text{ T}$, $\Phi \approx 1.01\text{ mWb}$.

(c) Reluctance

$$\mathcal{R} = \frac{l}{\mu_0 \mu_r A} = \frac{0.5}{1.2566\times10^{-3} \times 5\times10^{-4}} = \frac{0.5}{6.283\times10^{-7}} = 7.96\times10^{5}\text{ AT/Wb}$$

Check: $\Phi = \mathcal{F}/\mathcal{R} = 800 / 7.96\times10^{5} = 1.005\times10^{-3}\text{ Wb}$. ✓

$\mathcal{R} = 7.96\times10^{5}\text{ AT/Wb}$.

(a) Peak, RMS and average values

Peak value: $V_m = 325\text{ V}$.

RMS value: $V_{rms} = \dfrac{V_m}{\sqrt2} = \dfrac{325}{1.4142} = 229.8\text{ V} \approx 230\text{ V}$.

Average value (over half cycle): $V_{avg} = \dfrac{2 V_m}{\pi} = \dfrac{2 \times 325}{3.1416} = \dfrac{650}{3.1416} = 206.9\text{ V}$.

$V_m = 325\text{ V}$, $V_{rms} = 229.8\text{ V}$, $V_{avg} = 206.9\text{ V}$.

(b) Frequency and period

From $v = V_m\sin(\omega t)$, $\omega = 314\text{ rad/s}$.

Frequency: $f = \dfrac{\omega}{2\pi} = \dfrac{314}{6.2832} = 49.97\text{ Hz} \approx 50\text{ Hz}$.

Time period: $T = \dfrac{1}{f} = \dfrac{1}{50} = 0.02\text{ s} = 20\text{ ms}$.

$f \approx 50\text{ Hz}$, $T = 20\text{ ms}$.

(c) Form factor and peak factor

Form factor is the ratio of RMS value to average value:

$$k_f = \frac{V_{rms}}{V_{avg}} = \frac{229.8}{206.9} = 1.11$$

Peak (crest) factor is the ratio of peak value to RMS value:

$$k_p = \frac{V_m}{V_{rms}} = \frac{325}{229.8} = 1.414$$

These are the standard values for a pure sine wave ($1.11$ and $1.414$).

Form factor $= 1.11$, Peak factor $= 1.414$.

Given: $V = 220\text{ V}$, $R_a = 0.5\,\Omega$, $R_{sh} = 110\,\Omega$, line current $I_L = 24\text{ A}$.

(a) Field and armature currents

Field (shunt) current: $I_{sh} = \dfrac{V}{R_{sh}} = \dfrac{220}{110} = 2\text{ A}$.

Armature current: $I_a = I_L - I_{sh} = 24 - 2 = 22\text{ A}$.

$I_{sh} = 2\text{ A}$, $I_a = 22\text{ A}$.

(b) Back emf

For a motor: $E_b = V - I_a R_a = 220 - (22 \times 0.5) = 220 - 11 = 209\text{ V}$.

$E_b = 209\text{ V}$.

(c) Mechanical power developed

The mechanical power developed in the armature equals the back-emf times the armature current:

$$P_{mech} = E_b I_a = 209 \times 22 = 4598\text{ W} \approx 4.6\text{ kW}$$

(Cross-check: armature input $= V I_a = 220\times22 = 4840\text{ W}$; armature copper loss $= I_a^2 R_a = 22^2 \times 0.5 = 242\text{ W}$; developed $= 4840 - 242 = 4598\text{ W}$. ✓)

$P_{mech} = 4598\text{ W} \approx 4.6\text{ kW}$.

(a) Kirchhoff's laws

KCL (Current Law): The algebraic sum of currents meeting at any node (junction) of an electric circuit is zero — i.e. the total current entering a node equals the total current leaving it. It expresses conservation of charge.

KVL (Voltage Law): The algebraic sum of all voltages (emfs and IR drops) around any closed loop of a circuit is zero. It expresses conservation of energy.

(b) Parallel resistors across 60 V

Since all resistors are directly across the source, each has the full $60\text{ V}$:

$$I_1 = \frac{V}{R_1} = \frac{60}{10} = 6\text{ A}$$ $$I_2 = \frac{V}{R_2} = \frac{60}{20} = 3\text{ A}$$ $$I_3 = \frac{V}{R_3} = \frac{60}{30} = 2\text{ A}$$

Total current (by KCL): $I_T = I_1 + I_2 + I_3 = 6 + 3 + 2 = 11\text{ A}$.

Equivalent resistance check: $\dfrac{1}{R_{eq}} = \dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{30} = \dfrac{6+3+2}{60} = \dfrac{11}{60}$, so $R_{eq} = \dfrac{60}{11} = 5.45\,\Omega$, and $I_T = 60/5.45 = 11\text{ A}$. ✓

Total power supplied: $P = V I_T = 60 \times 11 = 660\text{ W}$.

$I_1 = 6\text{ A}$, $I_2 = 3\text{ A}$, $I_3 = 2\text{ A}$, $I_T = 11\text{ A}$, $P = 660\text{ W}$.

(a) Power triangle of the load

Apparent power: $S = V I = 230 \times 40 = 9200\text{ VA} = 9.2\text{ kVA}$.

Active power: $P = S\cos\phi = 9200 \times 0.6 = 5520\text{ W} = 5.52\text{ kW}$.

Power factor angle: $\phi_1 = \cos^{-1}(0.6) = 53.13^\circ$, $\sin\phi_1 = 0.8$.

Reactive power: $Q_1 = S\sin\phi_1 = 9200 \times 0.8 = 7360\text{ VAR} = 7.36\text{ kVAR}$.

$P = 5.52\text{ kW}$, $Q_1 = 7.36\text{ kVAR}$, $S = 9.2\text{ kVA}$.

(b) Capacitor for pf correction to 0.9 lagging

The active power $P$ stays constant at $5520\text{ W}$. New angle:

$$\phi_2 = \cos^{-1}(0.9) = 25.84^\circ,\quad \tan\phi_2 = 0.4843$$

New reactive power required from the source:

$$Q_2 = P\tan\phi_2 = 5520 \times 0.4843 = 2673\text{ VAR}$$

Reactive power the capacitor must supply:

$$Q_C = Q_1 - Q_2 = 7360 - 2673 = 4687\text{ VAR} \approx 4.69\text{ kVAR}$$

(Equivalently $Q_C = P(\tan\phi_1 - \tan\phi_2) = 5520(1.3333 - 0.4843) = 5520 \times 0.8490 = 4687\text{ VAR}$.)

$Q_C \approx 4.69\text{ kVAR}$.