BE Civil Engineering (IOE, TU) Basic Electrical Engineering (IOE, EE 401) Question Paper 2080 Nepal

Mesh Analysis

Define two clockwise mesh currents: $I_1$ in the left loop ($E_1, R_1, R_2$) and $I_2$ in the right loop ($R_2, R_3, E_2$). The middle resistor $R_2$ is shared.

Left mesh (KVL):

$$E_1 = I_1 R_1 + (I_1 - I_2)R_2$$ $$24 = 4I_1 + 12(I_1 - I_2) = 16I_1 - 12I_2 \quad (1)$$

Right mesh (KVL): Taking $E_2$ driving current into the node (source polarity aids current up through $R_2$ from right loop):

$$-E_2 = I_2 R_3 + (I_2 - I_1)R_2$$ $$-12 = 6I_2 + 12(I_2 - I_1) = -12I_1 + 18I_2 \quad (2)$$

From (1): $16I_1 - 12I_2 = 24$. From (2): $-12I_1 + 18I_2 = -12$.

Multiply (1) by 3 and (2) by 2:

$$48I_1 - 36I_2 = 72$$ $$-24I_1 + 36I_2 = -24$$

Add: $24I_1 = 48 \Rightarrow I_1 = 2\text{ A}$.

Substitute into (1): $16(2) - 12I_2 = 24 \Rightarrow 32 - 12I_2 = 24 \Rightarrow I_2 = \tfrac{8}{12} = 0.667\text{ A}$.

Current through $R_2$:

$$I_{R_2} = I_1 - I_2 = 2 - 0.667 = 1.333\text{ A (downward)}$$

Power in $R_2$:

$$P_{R_2} = I_{R_2}^2 R_2 = (1.333)^2 \times 12 = 1.778 \times 12 = \mathbf{21.33\ W}$$

Verification by Superposition

Source $E_1$ alone (short $E_2$): $R_3$ is in parallel with $R_2$.

$$R_{2\parallel3} = \frac{12 \times 6}{12+6} = 4\,\Omega,\quad R_{tot} = R_1 + 4 = 8\,\Omega$$ $$I_{src} = 24/8 = 3\text{ A},\quad V_{node} = 3 \times 4 = 12\text{ V}$$ $$I'_{R_2} = 12/12 = 1\text{ A (down)}$$

Source $E_2$ alone (short $E_1$): $R_1$ in parallel with $R_2$.

$$R_{1\parallel2} = \frac{4 \times 12}{4+12} = 3\,\Omega,\quad R_{tot} = R_3 + 3 = 9\,\Omega$$ $$I_{src} = 12/9 = 1.333\text{ A},\quad V_{node} = 1.333 \times 3 = 4\text{ V}$$ $$I''_{R_2} = 4/12 = 0.333\text{ A (down)}$$

Total: $I_{R_2} = 1 + 0.333 = \mathbf{1.333\ A}$, matching the mesh result. $\checkmark$

(a) Thevenin Equivalent

Remove $R_L$. The open-circuit voltage across the node-to-ground terminals is the voltage across $R_2$ (voltage divider of the $48\text{ V}$ source between $R_1$ and $R_2$):

$$V_{Th} = 48 \times \frac{R_2}{R_1 + R_2} = 48 \times \frac{24}{8+24} = 48 \times \frac{24}{32} = \mathbf{36\ V}$$

For $R_{Th}$, deactivate the source (short the $48\text{ V}$). Then $R_1$ and $R_2$ are in parallel as seen from the terminals:

$$R_{Th} = R_1 \parallel R_2 = \frac{8 \times 24}{8+24} = \frac{192}{32} = \mathbf{6\ \Omega}$$

(b) Maximum Power Transfer

Maximum power is delivered when:

$$R_L = R_{Th} = \mathbf{6\ \Omega}$$

The load current is:

$$I_L = \frac{V_{Th}}{R_{Th}+R_L} = \frac{36}{6+6} = 3\text{ A}$$

Maximum power in the load:

$$P_{max} = I_L^2 R_L = (3)^2 \times 6 = 9 \times 6 = \mathbf{54\ W}$$

Equivalently $P_{max} = \dfrac{V_{Th}^2}{4R_{Th}} = \dfrac{36^2}{4 \times 6} = \dfrac{1296}{24} = 54\text{ W}$. $\checkmark$

(c) Statement & Justification

Maximum Power Transfer Theorem: A resistive load receives maximum power from a source network when the load resistance equals the Thevenin resistance of the network, $R_L = R_{Th}$.

Justification: With $P_L = \dfrac{V_{Th}^2 R_L}{(R_{Th}+R_L)^2}$, setting $\dfrac{dP_L}{dR_L}=0$ gives $(R_{Th}+R_L)^2 - R_L \cdot 2(R_{Th}+R_L)=0 \Rightarrow R_{Th}+R_L = 2R_L \Rightarrow R_L = R_{Th}$. At this point efficiency is only 50% since equal power is lost in $R_{Th}$.

(a) Impedance, Current, Power Factor, Power at 50 Hz

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

Inductive reactance:

$$X_L = \omega L = 314.16 \times 0.1 = 31.42\,\Omega$$

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: $X = X_L - X_C = 31.42 - 31.83 = -0.41\,\Omega$ (slightly capacitive).

Impedance:

$$Z = \sqrt{R^2 + X^2} = \sqrt{10^2 + (-0.41)^2} = \sqrt{100 + 0.17} = \mathbf{10.008\ \Omega}$$

Current:

$$I = \frac{V}{Z} = \frac{230}{10.008} = \mathbf{22.98\ A}$$

Power factor:

$$\cos\phi = \frac{R}{Z} = \frac{10}{10.008} = \mathbf{0.9992\ (leading)}$$

Real power:

$$P = VI\cos\phi = 230 \times 22.98 \times 0.9992 = \mathbf{5281\ W \approx 5.28\ kW}$$

(Check: $P = I^2R = 22.98^2 \times 10 = 5281\text{ W}$. $\checkmark$)

(b) Resonant Frequency

$$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 100\times10^{-6}}} = \frac{1}{2\pi\sqrt{1\times10^{-5}}}$$ $$= \frac{1}{2\pi \times 3.162\times10^{-3}} = \frac{1}{0.01987} = \mathbf{50.33\ Hz}$$

(c) Quality Factor and Resonant Current

At resonance $\omega_0 = 2\pi f_0 = 316.2\text{ rad/s}$:

$$Q = \frac{\omega_0 L}{R} = \frac{316.2 \times 0.1}{10} = \frac{31.62}{10} = \mathbf{3.16}$$

Equivalently $Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} = \dfrac{1}{10}\sqrt{\dfrac{0.1}{100\times10^{-6}}} = \dfrac{1}{10}\sqrt{1000} = \dfrac{31.62}{10} = 3.16$. $\checkmark$

At resonance $Z = R = 10\,\Omega$, so the current is maximum:

$$I_0 = \frac{V}{R} = \frac{230}{10} = \mathbf{23\ A}$$

(a) Phase Voltage, Current, Power Factor (Star)

Phase voltage (star):

$$V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400}{1.732} = 230.9\text{ V}$$

Per-phase impedance:

$$Z_{ph} = \sqrt{R^2 + X_L^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10\,\Omega$$

Phase current (= line current in star):

$$I_{ph} = I_L = \frac{V_{ph}}{Z_{ph}} = \frac{230.9}{10} = \mathbf{23.09\ A}$$

Power factor:

$$\cos\phi = \frac{R}{Z} = \frac{8}{10} = \mathbf{0.8\ (lagging)}$$

(b) Three-Phase Powers (Star)

Real power:

$$P = \sqrt{3}\,V_L I_L \cos\phi = 1.732 \times 400 \times 23.09 \times 0.8 = \mathbf{12{,}800\ W \approx 12.8\ kW}$$

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

Reactive power ($\sin\phi = 0.6$):

$$Q = \sqrt{3}\,V_L I_L \sin\phi = 1.732 \times 400 \times 23.09 \times 0.6 = \mathbf{9600\ VAR}$$

Apparent power:

$$S = \sqrt{3}\,V_L I_L = 1.732 \times 400 \times 23.09 = \mathbf{16{,}000\ VA = 16\ kVA}$$

(Check: $S = \sqrt{P^2+Q^2} = \sqrt{12800^2 + 9600^2} = \sqrt{16000^2} = 16000\text{ VA}$. $\checkmark$)

(c) Delta Reconnection

In delta the phase voltage equals the line voltage:

$$V_{ph,\Delta} = V_L = 400\text{ V}$$

Phase current:

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

Line current in delta:

$$I_{L,\Delta} = \sqrt{3}\,I_{ph,\Delta} = 1.732 \times 40 = \mathbf{69.28\ A}$$

Note this is exactly three times the star line current ($3 \times 23.09 = 69.28\text{ A}$), confirming that delta draws three times the power of star for the same impedances.

(a) Efficiency at Full and Half Load (0.8 pf)

Output at full load: $S \times \text{pf} = 25{,}000 \times 0.8 = 20{,}000\text{ W}$.

Full load (copper loss = 400 W, iron loss = 250 W):

$$\eta_{FL} = \frac{20{,}000}{20{,}000 + 400 + 250} = \frac{20{,}000}{20{,}650} = 0.9685 = \mathbf{96.85\%}$$

Half load: output $= 0.5 \times 20{,}000 = 10{,}000\text{ W}$. Copper loss scales with (load fraction)$^2$: $P_{cu} = 0.5^2 \times 400 = 100\text{ W}$. Iron loss constant = 250 W.

$$\eta_{HL} = \frac{10{,}000}{10{,}000 + 100 + 250} = \frac{10{,}000}{10{,}350} = 0.9662 = \mathbf{96.62\%}$$

(b) Maximum Efficiency

Maximum efficiency occurs when copper loss = iron loss. Let $x$ = fraction of full load:

$$x^2 P_{cu,FL} = P_{iron} \Rightarrow x = \sqrt{\frac{250}{400}} = \sqrt{0.625} = 0.7906$$

So maximum efficiency occurs at $\mathbf{79.06\%}$ of full load.

Output at this load: $0.7906 \times 20{,}000 = 15{,}811\text{ W}$. At that point copper loss = iron loss = 250 W, total loss = 500 W.

$$\eta_{max} = \frac{15{,}811}{15{,}811 + 500} = \frac{15{,}811}{16{,}311} = 0.9693 = \mathbf{96.93\%}$$

(c) Turns Ratio and Secondary Current

Turns ratio:

$$\frac{N_1}{N_2} = \frac{V_1}{V_2} = \frac{2000}{200} = \mathbf{10:1}$$

Full-load secondary current:

$$I_2 = \frac{S}{V_2} = \frac{25{,}000}{200} = \mathbf{125\ A}$$

(Primary full-load current $I_1 = 25{,}000/2000 = 12.5\text{ A}$.)

(a) Equivalent Capacitance

Parallel combination of $C_2$ and $C_3$:

$$C_{23} = C_2 + C_3 = 6 + 12 = 18\,\mu\text{F}$$

$C_1$ in series with $C_{23}$:

$$C_{eq} = \frac{C_1 \times C_{23}}{C_1 + C_{23}} = \frac{4 \times 18}{4 + 18} = \frac{72}{22} = \mathbf{3.27\ \mu F}$$

(b) Total Charge and Energy

Charge drawn from supply (series elements carry the same charge):

$$Q = C_{eq} V = 3.27\times10^{-6} \times 100 = \mathbf{327\ \mu C}$$

Energy stored:

$$W = \tfrac{1}{2}C_{eq}V^2 = \tfrac{1}{2}\times 3.27\times10^{-6}\times 100^2 = \tfrac{1}{2}\times 3.27\times10^{-6}\times 10^4 = \mathbf{16.36\ mJ}$$

(c) Voltage across $C_1$

$C_1$ carries the full series charge $Q = 327\,\mu\text{C}$:

$$V_1 = \frac{Q}{C_1} = \frac{327\times10^{-6}}{4\times10^{-6}} = \mathbf{81.8\ V}$$

(Check: voltage across the parallel section $V_{23} = 100 - 81.8 = 18.2\text{ V}$; charge there $= 18\,\mu F \times 18.2\text{ V} = 327\,\mu C$. $\checkmark$)

(a) Magnetomotive Force

$$\text{MMF} = N I = 500 \times 2 = \mathbf{1000\ At\ (ampere-turns)}$$

(b) Reluctance

$$S = \frac{l}{\mu_0 \mu_r A} = \frac{0.4}{(4\pi\times10^{-7})(800)(4\times10^{-4})}$$

Denominator: $4\pi\times10^{-7} = 1.2566\times10^{-6}$; times $800 = 1.0053\times10^{-3}$; times $4\times10^{-4} = 4.021\times10^{-7}$.

$$S = \frac{0.4}{4.021\times10^{-7}} = \mathbf{9.95\times10^{5}\ At/Wb}$$

(c) Flux and Flux Density

Magnetic flux:

$$\Phi = \frac{\text{MMF}}{S} = \frac{1000}{9.95\times10^{5}} = 1.005\times10^{-3}\text{ Wb} = \mathbf{1.005\ mWb}$$

Flux density:

$$B = \frac{\Phi}{A} = \frac{1.005\times10^{-3}}{4\times10^{-4}} = \mathbf{2.51\ T}$$

(Check via $B = \mu_0\mu_r H$, with $H = NI/l = 1000/0.4 = 2500\text{ At/m}$: $B = 1.2566\times10^{-6}\times800\times2500 = 2.51\text{ T}$. $\checkmark$)

(a) Peak, RMS, Average, Frequency

From $v(t) = V_m\sin(\omega t)$: peak $V_m = \mathbf{311\ V}$, $\omega = 314\text{ rad/s}$.

RMS value:

$$V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{311}{1.414} = \mathbf{219.9\ V \approx 220\ V}$$

Average over half-cycle:

$$V_{avg} = \frac{2V_m}{\pi} = \frac{2\times311}{3.1416} = \mathbf{197.9\ V}$$

Frequency:

$$f = \frac{\omega}{2\pi} = \frac{314}{6.2832} = \mathbf{49.97\ Hz \approx 50\ Hz}$$

(b) Form Factor and Peak Factor

Form factor = RMS / average:

$$k_f = \frac{V_{rms}}{V_{avg}} = \frac{219.9}{197.9} = \mathbf{1.11}$$

Peak (crest) factor = peak / RMS:

$$k_p = \frac{V_m}{V_{rms}} = \frac{311}{219.9} = \mathbf{1.414}$$

(c) Instantaneous Value at t = 2.5 ms

$$\omega t = 314 \times 2.5\times10^{-3} = 0.785\text{ rad} \;(=45^\circ)$$ $$v = 311\sin(0.785) = 311 \times 0.7071 = \mathbf{219.9\ V}$$

(a) Field and Armature Currents

Field (shunt) current:

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

Armature current (line current minus field current):

$$I_a = I_L - I_{sh} = 42 - 2 = \mathbf{40\ A}$$

(b) Back EMF

For a motor, $E_b = V - I_a R_a$:

$$E_b = 220 - (40 \times 0.5) = 220 - 20 = \mathbf{200\ V}$$

(c) Gross Mechanical Power Developed

The power developed in the armature equals the back EMF times armature current:

$$P_{dev} = E_b I_a = 200 \times 40 = 8000\text{ W} = \mathbf{8\ kW}$$

(For reference, electrical input $= V I_L = 220\times42 = 9240\text{ W}$; armature copper loss $= I_a^2 R_a = 40^2\times0.5 = 800\text{ W}$; field loss $= 220\times2 = 440\text{ W}$; sum of losses $+ P_{dev} = 800+440+8000 = 9240\text{ W}$. $\checkmark$)

(a) Total Resistance and Current

Parallel conductance:

$$\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} = \frac{4}{20} + \frac{2}{20} + \frac{1}{20} = \frac{7}{20}$$ $$R_{eq} = \frac{20}{7} = \mathbf{2.857\ \Omega}$$

Total current:

$$I = \frac{V}{R_{eq}} = \frac{60}{2.857} = \mathbf{21\ A}$$

(b) Branch Currents

$$I_1 = \frac{60}{5} = 12\text{ A},\quad I_2 = \frac{60}{10} = 6\text{ A},\quad I_3 = \frac{60}{20} = 3\text{ A}$$

Sum $= 12 + 6 + 3 = 21\text{ A}$, matching total current (KCL). $\checkmark$

(c) Power Dissipation

Total power:

$$P = VI = 60 \times 21 = \mathbf{1260\ W}$$

Individual powers:

$$P_1 = 60\times12 = 720\text{ W},\; P_2 = 60\times6 = 360\text{ W},\; P_3 = 60\times3 = 180\text{ W}$$ $$P_1 + P_2 + P_3 = 720 + 360 + 180 = 1260\text{ W} = P. \;\checkmark$$

(a) Power Triangle

Apparent power:

$$S = VI = 230 \times 15 = \mathbf{3450\ VA}$$

Real power ($\cos\phi_1 = 0.7$):

$$P = S\cos\phi_1 = 3450 \times 0.7 = \mathbf{2415\ W}$$

Reactive power ($\sin\phi_1 = \sqrt{1-0.7^2} = \sqrt{0.51} = 0.714$):

$$Q_1 = S\sin\phi_1 = 3450 \times 0.714 = \mathbf{2464\ VAR\ (lagging)}$$

(b) Capacitor for Power-Factor Correction

Real power stays constant at $P = 2415\text{ W}$. New angle: $\cos\phi_2 = 0.95 \Rightarrow \phi_2 = 18.19^\circ$, $\tan\phi_2 = 0.3287$.

New reactive power:

$$Q_2 = P\tan\phi_2 = 2415 \times 0.3287 = 793.8\text{ VAR}$$

Reactive power supplied by capacitor:

$$Q_C = Q_1 - Q_2 = 2464 - 793.8 = 1670\text{ VAR}$$

Capacitor reactive power $Q_C = \dfrac{V^2}{X_C} = V^2\omega C$, so:

$$C = \frac{Q_C}{2\pi f V^2} = \frac{1670}{2\pi \times 50 \times 230^2} = \frac{1670}{314.16 \times 52900} = \frac{1670}{1.662\times10^{7}}$$ $$C = 1.005\times10^{-4}\text{ F} = \mathbf{100.5\ \mu F}$$