BE Civil Engineering (IOE, TU) Basic Electrical Engineering (IOE, EE 401) Question Paper 2076 Nepal

Step 1 — Assign mesh currents. Let $I_1$ (left loop) and $I_2$ (right loop) both circulate clockwise. The shared $5\,\Omega$ branch carries $(I_1-I_2)$ downward.

Step 2 — Write KVL for each mesh.

Left mesh: $20 = 2I_1 + 5(I_1-I_2) \Rightarrow 7I_1 - 5I_2 = 20$

Right mesh (10 V source drives $I_2$): $10 = 3I_2 + 5(I_2-I_1) \Rightarrow -5I_1 + 8I_2 = 10$

Step 3 — Solve the simultaneous equations.

$$\begin{cases} 7I_1 - 5I_2 = 20 \\ -5I_1 + 8I_2 = 10 \end{cases}$$

Determinant: $\Delta = (7)(8) - (-5)(-5) = 56 - 25 = 31$.

$$I_1 = \frac{\begin{vmatrix}20 & -5\\ 10 & 8\end{vmatrix}}{31} = \frac{(20)(8)-(-5)(10)}{31} = \frac{160+50}{31} = \frac{210}{31} = 6.774\,\text{A}$$ $$I_2 = \frac{\begin{vmatrix}7 & 20\\ -5 & 10\end{vmatrix}}{31} = \frac{(7)(10)-(20)(-5)}{31} = \frac{70+100}{31} = \frac{170}{31} = 5.484\,\text{A}$$

Step 4 — Current through the 5 Ω resistor.

$$I_{5\Omega} = I_1 - I_2 = 6.774 - 5.484 = 1.290\,\text{A (downward)}$$

Step 5 — Power dissipated.

$$P = I_{5\Omega}^2 \times R = (1.290)^2 \times 5 = 1.664 \times 5 = 8.32\,\text{W}$$

Final answers: $\mathbf{I_{5\Omega} = 1.29\ A}$ (flowing downward through the centre branch), dissipating $\mathbf{P \approx 8.32\ W}$.

Thevenin's theorem: Any linear two-terminal network of sources and resistances can be replaced by a single voltage source $V_{Th}$ in series with a single resistance $R_{Th}$, where $V_{Th}$ is the open-circuit voltage at the terminals and $R_{Th}$ is the resistance seen from the terminals with all independent sources replaced by their internal resistances (voltage sources short-circuited, current sources open-circuited).

Step 1 — Open-circuit voltage $V_{Th}$ (remove $R_L$).

With A-B open, current flows through $R_1$ and $R_2$ in series:

$$I = \frac{24}{6+12} = \frac{24}{18} = 1.333\,\text{A}$$ $$V_{Th} = V_{R_2} = I \times R_2 = 1.333 \times 12 = 16\,\text{V}$$

Step 2 — Thevenin resistance $R_{Th}$. Short the 24 V source. Then $R_1$ and $R_2$ are in parallel as seen from A-B:

$$R_{Th} = \frac{R_1 R_2}{R_1+R_2} = \frac{6\times 12}{6+12} = \frac{72}{18} = 4\,\Omega$$

Step 3 — Maximum power transfer. Maximum power is transferred when

$$\boxed{R_L = R_{Th} = 4\,\Omega}$$

Step 4 — Maximum power.

$$P_{max} = \frac{V_{Th}^2}{4R_{Th}} = \frac{16^2}{4\times 4} = \frac{256}{16} = 16\,\text{W}$$

Final answers: $V_{Th}=16\ \text{V}$, $R_{Th}=4\ \Omega$, $\mathbf{R_L = 4\ \Omega}$, $\mathbf{P_{max}=16\ W}$.

Given: $R=12\,\Omega$, $L=0.15\,\text{H}$, $C=100\,\mu\text{F}=100\times10^{-6}\,\text{F}$, $V=230\,\text{V}$, $f=50\,\text{Hz}$, $\omega = 2\pi f = 314.16\,\text{rad/s}$.

(a) Reactances, impedance, current, phase angle.

$$X_L = \omega L = 314.16 \times 0.15 = 47.12\,\Omega$$ $$X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times 100\times10^{-6}} = \frac{1}{0.031416} = 31.83\,\Omega$$ $$X = X_L - X_C = 47.12 - 31.83 = 15.29\,\Omega\ (\text{inductive})$$ $$Z = \sqrt{R^2 + X^2} = \sqrt{12^2 + 15.29^2} = \sqrt{144 + 233.8} = \sqrt{377.8} = 19.44\,\Omega$$ $$I = \frac{V}{Z} = \frac{230}{19.44} = 11.83\,\text{A}$$ $$\phi = \tan^{-1}\!\left(\frac{X}{R}\right) = \tan^{-1}\!\left(\frac{15.29}{12}\right) = \tan^{-1}(1.274) = 51.9^{\circ}\ (\text{current lags voltage})$$

Power factor: $\cos\phi = R/Z = 12/19.44 = 0.617$ lagging.

(b) Powers.

$$P = VI\cos\phi = 230\times11.83\times0.617 = 1679\,\text{W} \approx 1.68\,\text{kW}$$ $$Q = VI\sin\phi = 230\times11.83\times\sin51.9^{\circ} = 230\times11.83\times0.787 = 2141\,\text{VAR} \approx 2.14\,\text{kVAR}$$ $$S = VI = 230\times11.83 = 2721\,\text{VA} \approx 2.72\,\text{kVA}$$

(Check: $S=\sqrt{P^2+Q^2}=\sqrt{1679^2+2141^2}=\sqrt{2.82\times10^6+4.58\times10^6}=2720\,\text{VA}$ ✓)

(c) Resonant frequency.

$$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.15\times100\times10^{-6}}} = \frac{1}{2\pi\sqrt{1.5\times10^{-5}}}$$ $$= \frac{1}{2\pi\times 3.873\times10^{-3}} = \frac{1}{0.02433} = 41.1\,\text{Hz}$$

Final answers: $Z=19.44\ \Omega$, $\mathbf{I=11.83\ A}$, $\phi=51.9^{\circ}$ lagging; $P\approx1.68\ \text{kW}$, $Q\approx2.14\ \text{kVAR}$, $S\approx2.72\ \text{kVA}$; $\mathbf{f_0\approx41.1\ Hz}$.

Given: Star (Y) connection, $V_L = 400\,\text{V}$, per-phase $R=15\,\Omega$, $X_L=20\,\Omega$.

(a) Phase voltage. In star, $V_{ph} = \dfrac{V_L}{\sqrt3}$:

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

(b) Phase impedance and current.

$$Z_{ph} = \sqrt{R^2 + X_L^2} = \sqrt{15^2 + 20^2} = \sqrt{225+400} = \sqrt{625} = 25\,\Omega$$ $$I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{230.9}{25} = 9.24\,\text{A}$$

In star, line current = phase current, so $I_L = 9.24\,\text{A}$.

(c) Power factor.

$$\cos\phi = \frac{R}{Z_{ph}} = \frac{15}{25} = 0.6\ \text{lagging}$$

(d) Total active power.

$$P = \sqrt3\,V_L I_L \cos\phi = 1.732\times400\times9.24\times0.6 = 3841\,\text{W} \approx 3.84\,\text{kW}$$

(Check via per-phase: $P = 3 I_{ph}^2 R = 3\times9.24^2\times15 = 3\times85.4\times15 = 3842\,\text{W}$ ✓)

Final answers: $V_{ph}=230.9\ \text{V}$, $\mathbf{I_L=9.24\ A}$, $\text{p.f.}=0.6$ lagging, $\mathbf{P\approx3.84\ kW}$.

Given: $S=25\,\text{kVA}$, iron loss $P_i = 250\,\text{W}$ (constant), full-load copper loss $P_{cu,FL}=350\,\text{W}$.

(a) Efficiency at 0.8 p.f. lagging.

Output at load fraction $x$: $P_{out} = x \cdot S \cdot \cos\phi$. Copper loss varies as $x^2$.

$$\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2 P_{cu,FL}}$$

Full load ($x=1$): output $= 25000\times0.8 = 20000\,\text{W}$; losses $=250+350=600\,\text{W}$.

$$\eta_{FL} = \frac{20000}{20000+600} = \frac{20000}{20600} = 0.9709 = 97.09\%$$

Half load ($x=0.5$): output $=0.5\times25000\times0.8=10000\,\text{W}$; copper loss $=(0.5)^2\times350=87.5\,\text{W}$; total losses $=250+87.5=337.5\,\text{W}$.

$$\eta_{HL} = \frac{10000}{10000+337.5} = \frac{10000}{10337.5} = 0.9674 = 96.74\%$$

(b) Maximum-efficiency load. Maximum efficiency occurs when variable (copper) loss equals constant (iron) loss: $x^2 P_{cu,FL} = P_i$.

$$x = \sqrt{\frac{P_i}{P_{cu,FL}}} = \sqrt{\frac{250}{350}} = \sqrt{0.7143} = 0.845$$

So maximum efficiency occurs at $0.845\times25 = 21.1\,\text{kVA}$.

At this load (unity p.f.), copper loss $=$ iron loss $=250\,\text{W}$, total losses $=500\,\text{W}$; output $=0.845\times25000\times1.0 = 21125\,\text{W}$.

$$\eta_{max} = \frac{21125}{21125+500} = \frac{21125}{21625} = 0.9769 = 97.69\%$$

Final answers: $\eta_{FL}=\mathbf{97.09\%}$, $\eta_{HL}=96.74\%$; max efficiency at $\mathbf{x=0.845}$ (≈21.1 kVA) giving $\eta_{max}=\mathbf{97.69\%}$ at unity p.f.

Given: $C_1=4\,\mu\text{F}$, $C_2=6\,\mu\text{F}$, $C_3=12\,\mu\text{F}$, $V=300\,\text{V}$, series connection.

(a) Equivalent capacitance.

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

(b) Total charge. In series, every capacitor carries the same charge equal to the total charge:

$$Q = C_{eq}\,V = 2\times10^{-6}\times300 = 600\times10^{-6}\,\text{C} = 600\,\mu\text{C}$$

(c) Voltage across each capacitor ($V=Q/C$):

$$V_1 = \frac{600\,\mu\text{C}}{4\,\mu\text{F}} = 150\,\text{V}$$ $$V_2 = \frac{600}{6} = 100\,\text{V}$$ $$V_3 = \frac{600}{12} = 50\,\text{V}$$

(Check: $150+100+50 = 300\,\text{V}$ ✓)

Final answers: $C_{eq}=\mathbf{2\ \mu F}$, $Q=\mathbf{600\ \mu C}$, $V_1=150\ \text{V}$, $V_2=100\ \text{V}$, $V_3=50\ \text{V}$.

Given: $l = 40\,\text{cm} = 0.40\,\text{m}$, $A = 5\,\text{cm}^2 = 5\times10^{-4}\,\text{m}^2$, $N=200$, $I=2\,\text{A}$, $\mu_r=800$.

(a) MMF.

$$\mathcal{F} = NI = 200\times2 = 400\,\text{A-t (ampere-turns)}$$

(b) Magnetic field strength.

$$H = \frac{NI}{l} = \frac{400}{0.40} = 1000\,\text{A/m}$$

(c) Flux density.

$$B = \mu_0\mu_r H = (4\pi\times10^{-7})(800)(1000)$$ $$= 1.2566\times10^{-6}\times 800\times1000 = 1.2566\times10^{-6}\times 8\times10^{5} = 1.005\,\text{T}$$

(d) Total flux.

$$\Phi = B \times A = 1.005 \times 5\times10^{-4} = 5.03\times10^{-4}\,\text{Wb} = 0.503\,\text{mWb}$$

Final answers: $\mathcal{F}=\mathbf{400\ A\text{-}t}$, $H=1000\ \text{A/m}$, $B\approx\mathbf{1.005\ T}$, $\Phi\approx\mathbf{0.503\ mWb}$.

Superposition: Find the contribution of each source acting alone (other voltage sources shorted), then add.

Source 1 — only 12 V active (6 V shorted). The $4\,\Omega$ centre branch is in parallel with $R_2=3\,\Omega$ (which is now connected to a short):

$$R_{4\parallel3} = \frac{4\times3}{4+3} = \frac{12}{7} = 1.714\,\Omega$$

Total resistance seen by 12 V source: $R_1 + 1.714 = 2 + 1.714 = 3.714\,\Omega$. Source current: $I_s = 12/3.714 = 3.231\,\text{A}$. This divides between the $4\,\Omega$ and $3\,\Omega$ branches; current in $4\,\Omega$ (current divider):

$$I_{4}' = I_s\times\frac{3}{4+3} = 3.231\times\frac{3}{7} = 1.385\,\text{A (downward)}$$

Source 2 — only 6 V active (12 V shorted). Now $4\,\Omega$ is in parallel with $R_1=2\,\Omega$:

$$R_{4\parallel2} = \frac{4\times2}{4+2} = \frac{8}{6} = 1.333\,\Omega$$

Total seen by 6 V source: $R_2 + 1.333 = 3 + 1.333 = 4.333\,\Omega$. Source current: $I_s' = 6/4.333 = 1.385\,\text{A}$. Current in $4\,\Omega$ branch (divider):

$$I_4'' = 1.385\times\frac{2}{4+2} = 1.385\times\frac{2}{6} = 0.462\,\text{A (downward)}$$

Total. Both contributions drive current downward through $R_3$, so they add:

$$I_4 = I_4' + I_4'' = 1.385 + 0.462 = 1.847\,\text{A}$$

Final answer: $\mathbf{I_{4\Omega} \approx 1.85\ A}$ (downward through the centre branch).

Definitions. The RMS (root-mean-square) value of an alternating quantity is the equivalent DC value that produces the same heating effect in a given resistance over one cycle. The average value (over a half cycle for a symmetrical AC wave) is the mean of all instantaneous values over that half cycle.

Given: $i(t)=14.14\sin(314t)\,\text{A}$, so peak $I_m=14.14\,\text{A}$ and $\omega = 314\,\text{rad/s}$.

(a) Peak value: $I_m = 14.14\,\text{A}$.

(b) RMS value:

$$I_{rms} = \frac{I_m}{\sqrt2} = \frac{14.14}{1.414} = 10.0\,\text{A}$$

(c) Average value (half cycle):

$$I_{avg} = \frac{2I_m}{\pi} = 0.637\times14.14 = 9.0\,\text{A}$$

(d) Form factor:

$$k_f = \frac{I_{rms}}{I_{avg}} = \frac{10.0}{9.0} = 1.11$$

(e) Frequency:

$$f = \frac{\omega}{2\pi} = \frac{314}{2\pi} = \frac{314}{6.283} = 50\,\text{Hz}$$

Final answers: $I_m=14.14\ \text{A}$, $\mathbf{I_{rms}=10\ A}$, $I_{avg}=9.0\ \text{A}$, form factor $=1.11$, $\mathbf{f=50\ Hz}$.

Given: $V=230\,\text{V}$, $R_a=0.5\,\Omega$, $R_{sh}=115\,\Omega$, line current $I_L=30\,\text{A}$ (shunt connection).

(a) Field (shunt) current.

$$I_{sh} = \frac{V}{R_{sh}} = \frac{230}{115} = 2\,\text{A}$$

(b) Armature current.

$$I_a = I_L - I_{sh} = 30 - 2 = 28\,\text{A}$$

(c) Back EMF. For a motor, $E_b = V - I_a R_a$:

$$E_b = 230 - 28\times0.5 = 230 - 14 = 216\,\text{V}$$

(d) Gross mechanical power developed in the armature.

$$P_m = E_b\,I_a = 216\times28 = 6048\,\text{W} \approx 6.05\,\text{kW}$$

Final answers: $I_{sh}=2\ \text{A}$, $I_a=28\ \text{A}$, $\mathbf{E_b=216\ V}$, $\mathbf{P_m\approx6.05\ kW}$.

Given: $P=5\,\text{kW}$, initial $\cos\phi_1=0.6$ lagging, target $\cos\phi_2=0.9$ lagging, $V=230\,\text{V}$, $f=50\,\text{Hz}$.

Phase angles. $\phi_1 = \cos^{-1}(0.6) = 53.13^{\circ}$, $\tan\phi_1 = 1.333$. $\phi_2 = \cos^{-1}(0.9) = 25.84^{\circ}$, $\tan\phi_2 = 0.4843$.

(a) Original reactive power.

$$Q_1 = P\tan\phi_1 = 5\times1.333 = 6.667\,\text{kVAR (lagging)}$$

(b) Reactive power after correction.

$$Q_2 = P\tan\phi_2 = 5\times0.4843 = 2.422\,\text{kVAR (lagging)}$$

(c) Capacitor rating (reactive power it must supply):

$$Q_C = Q_1 - Q_2 = 6.667 - 2.422 = 4.245\,\text{kVAR}$$

(d) Capacitance. The capacitor supplies $Q_C = \dfrac{V^2}{X_C} = V^2\omega C$, so

$$C = \frac{Q_C}{\omega V^2} = \frac{4245}{2\pi\times50\times230^2} = \frac{4245}{314.16\times52900}$$ $$= \frac{4245}{1.662\times10^{7}} = 2.554\times10^{-4}\,\text{F} = 255.4\,\mu\text{F}$$

Final answers: $Q_1=6.67\ \text{kVAR}$, $Q_2=2.42\ \text{kVAR}$, $\mathbf{Q_C\approx4.25\ kVAR}$, $\mathbf{C\approx255\ \mu F}$.