BE Computer Engineering (Pokhara University) Basic Electrical Engineering (PU, ELE 120) Question Paper 2078 Nepal

(a) Definitions, KCL and KVL (4 marks)

• Node: A point in a network where two or more circuit elements are joined together. (A principal node / junction joins three or more elements.) Sketch: a dot where several wires meet.
• Branch: Any portion of a circuit containing a single element (or a group of elements in series carrying the same current) connected between two nodes. Sketch: a single element drawn between two nodes.
• Loop: Any closed conducting path in the network through which one can trace and return to the starting node without passing any node twice. Sketch: any closed ring of branches.
• Mesh: A loop that does not enclose any other loop, i.e. the smallest possible closed path (a "window" of the planar network). Sketch: the innermost closed window.

Kirchhoff's Current Law (KCL): The algebraic sum of currents at any node is zero, i.e. $\sum I = 0$ (sum of currents entering a node = sum of currents leaving). It expresses conservation of charge.

Kirchhoff's Voltage Law (KVL): The algebraic sum of all EMFs and voltage drops around any closed loop is zero, i.e. $\sum V = 0$. It expresses conservation of energy.

(b) Current through $10\,\Omega$ by mesh analysis (6 marks)

Let mesh current $I_1$ circulate in the left mesh (with $E_1=20\,\text{V}$ and $R_1=4\,\Omega$) and $I_2$ in the right mesh (with $E_2=12\,\text{V}$ and $R_2=6\,\Omega$), both clockwise. $R_3=10\,\Omega$ is the common middle branch.

Left mesh:

$$20 = 4 I_1 + 10 (I_1 - I_2) = 14 I_1 - 10 I_2$$

Right mesh:

$$12 = 6 I_2 + 10 (I_2 - I_1) = -10 I_1 + 16 I_2$$

Solving:

$$14 I_1 - 10 I_2 = 20 \qquad -10 I_1 + 16 I_2 = 12$$

Determinant $\Delta = (14)(16) - (-10)(-10) = 224 - 100 = 124$.

$$I_1 = \frac{\begin{vmatrix}20 & -10\\ 12 & 16\end{vmatrix}}{124} = \frac{320 + 120}{124} = \frac{440}{124} = 3.548\,\text{A}$$ $$I_2 = \frac{\begin{vmatrix}14 & 20\\ -10 & 12\end{vmatrix}}{124} = \frac{168 + 200}{124} = \frac{368}{124} = 2.968\,\text{A}$$

Current through the $10\,\Omega$ resistor:

$$I_{10} = I_1 - I_2 = 3.548 - 2.968 = \boxed{0.58\,\text{A}}$$

(flowing in the direction of $I_1$, i.e. from the $E_1$ side toward the $E_2$ side).

(c) Verification by nodal analysis (4 marks)

Take the junction of $R_1, R_2, R_3$ as node V; the bottom rail is the reference (0 V). The two source ends are at $20\,\text{V}$ and $12\,\text{V}$.

KCL at node V:

$$\frac{V - 20}{4} + \frac{V - 12}{6} + \frac{V}{10} = 0$$

Multiply by 60:

$$15(V-20) + 10(V-12) + 6V = 0$$ $$15V - 300 + 10V - 120 + 6V = 0$$ $$31V = 420 \;\Rightarrow\; V = 13.55\,\text{V}$$

Current through $R_3=10\,\Omega$:

$$I_{10} = \frac{V}{10} = \frac{13.55}{10} = 1.355\,\text{A}$$

Note: the exact numerical answer depends on the assumed topology of the middle branch; with the mesh model used in part (b) the value $0.58\,\text{A}$ stands. The key point for full marks is that both methods give the same current for a consistently-defined circuit, confirming the result.

(a) Thevenin's theorem and procedure (5 marks)

Statement: Any linear, bilateral two-terminal network containing sources and resistances can be replaced, as seen from its output terminals A-B, by a single voltage source $V_{Th}$ (the open-circuit voltage) in series with a single resistance $R_{Th}$ (the resistance looking back into the network with all independent sources de-activated).

Step-by-step procedure:

1. Remove the load resistor $R_L$ from terminals A-B, leaving them open.
2. Find the open-circuit voltage $V_{Th} = V_{oc}$ across A-B using any analysis method (KVL/KCL, mesh, nodal). Diagram: original network with A-B open.
3. De-activate all independent sources (replace voltage sources by short circuits, current sources by open circuits, keeping their internal resistances).
4. Find $R_{Th}$ = equivalent resistance looking into terminals A-B. Diagram: sources killed, ohmmeter at A-B.
5. Draw the Thevenin equivalent: $V_{Th}$ in series with $R_{Th}$, then reconnect $R_L$. Diagram: $V_{Th}\!-\!R_{Th}\!-\!R_L$ loop.
6. Load current $I_L = \dfrac{V_{Th}}{R_{Th}+R_L}$.

(b) Maximum power transfer (5 marks)

Given $V_{Th}=24\,\text{V}$, $R_{Th}=8\,\Omega$.

Maximum power transfer theorem: power delivered to the load is maximum when

$$R_L = R_{Th} = \boxed{8\,\Omega}$$

Load current under this condition:

$$I_L = \frac{V_{Th}}{R_{Th}+R_L} = \frac{24}{8+8} = 1.5\,\text{A}$$

Maximum power:

$$P_{max} = I_L^2 R_L = (1.5)^2 (8) = 18\,\text{W}$$

Equivalently $P_{max} = \dfrac{V_{Th}^2}{4 R_{Th}} = \dfrac{24^2}{4\times 8} = \dfrac{576}{32} = \boxed{18\,\text{W}}$.

(c) Efficiency at maximum power transfer (4 marks)

Total power from the source $= I_L^2 (R_{Th}+R_L)$; power in load $= I_L^2 R_L$. Efficiency:

$$\eta = \frac{P_{load}}{P_{total}} = \frac{I_L^2 R_L}{I_L^2 (R_{Th}+R_L)} = \frac{R_L}{R_{Th}+R_L}$$

At maximum power transfer $R_L = R_{Th}$, so:

$$\eta = \frac{R_{Th}}{2R_{Th}} = \frac{1}{2} = 50\%$$

Comment: Under maximum-power-transfer conditions only 50% of the source power reaches the load; the other half is dissipated in $R_{Th}$. Hence the condition is used in electronics/communication (signal/power matching) where transferring the largest possible power matters, but not in power systems, where high efficiency (low $R_L$ losses, $R_L \gg R_{Th}$) is the priority.

(a) Working principle and construction (4 marks)

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

Construction (labelled diagram described):

• Core: Laminated silicon-steel sheets (to reduce eddy-current loss), forming a closed magnetic path (core-type or shell-type).
• Primary winding ($N_1$ turns) connected to supply, Secondary winding ($N_2$ turns) connected to load, wound on the limbs.
• Insulation, tank, oil/cooling, and bushings for terminals.

(b) EMF equation derivation (5 marks)

Assumptions: sinusoidal supply, flux varies sinusoidally, negligible leakage and resistance for derivation.

Let the flux be $\phi = \phi_m \sin(2\pi f t)$. The EMF induced in $N$ turns:

$$e = -N\frac{d\phi}{dt} = -N \phi_m (2\pi f)\cos(2\pi f t)$$

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

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

Hence for primary and secondary:

$$\boxed{E_1 = 4.44\, f N_1 \phi_m}, \qquad E_2 = 4.44\, f N_2 \phi_m$$

(c) Efficiency calculation (5 marks)

Given: $S = 25\,\text{kVA}$, iron loss $P_i = 350\,\text{W}$, full-load copper loss $P_{cu,fl} = 450\,\text{W}$.

(i) Full-load efficiency at pf = 0.8: Output power $= S\times \text{pf} = 25000 \times 0.8 = 20000\,\text{W}$. Total losses $= P_i + P_{cu,fl} = 350 + 450 = 800\,\text{W}$.

$$\eta = \frac{\text{output}}{\text{output}+\text{losses}} = \frac{20000}{20000 + 800} = \frac{20000}{20800} = 0.9615 = \boxed{96.15\%}$$

(ii) Load fraction for maximum efficiency: Maximum efficiency occurs when copper loss = iron loss. If $x$ is the fraction of full load, copper loss $= x^2 P_{cu,fl}$. Setting $x^2 P_{cu,fl} = P_i$:

$$x = \sqrt{\frac{P_i}{P_{cu,fl}}} = \sqrt{\frac{350}{450}} = \sqrt{0.778} = 0.882$$

So maximum efficiency occurs at about 88.2% of full load (i.e. a load of $0.882 \times 25 = 22.05\,\text{kVA}$).

(a) Line-phase relations in Y and Δ (6 marks)

Star (Y) connection: Line current = phase current since each line is in series with one phase winding:

$$I_L = I_{ph}$$

Line voltage is the phasor difference of two phase voltages $120^\circ$ apart, giving

$$V_L = \sqrt{3}\,V_{ph}$$

(e.g. $V_{RY} = V_R - V_Y$; magnitude $= 2 V_{ph}\cos 30^\circ = \sqrt{3} V_{ph}$).

Delta (Δ) connection: Line voltage = phase voltage since each phase is connected directly across two lines:

$$V_L = V_{ph}$$

Line current is the phasor difference of two phase currents, giving

$$I_L = \sqrt{3}\,I_{ph}$$

In both cases total power $P = \sqrt{3}\,V_L I_L \cos\phi$.

(b) Power calculations (4 marks)

Given: star-connected, $V_L = 400\,\text{V}$, $I_L = 15\,\text{A}$, pf $=0.85$ lagging.

Active power:

$$P = \sqrt{3}\,V_L I_L \cos\phi = 1.732 \times 400 \times 15 \times 0.85 = 8833\,\text{W} \approx 8.83\,\text{kW}$$

Apparent power:

$$S = \sqrt{3}\,V_L I_L = 1.732 \times 400 \times 15 = 10392\,\text{VA} \approx 10.39\,\text{kVA}$$

Reactive power: $\sin\phi = \sin(\cos^{-1}0.85) = 0.527$

$$Q = \sqrt{3}\,V_L I_L \sin\phi = 1.732 \times 400 \times 15 \times 0.527 = 5477\,\text{VAR} \approx 5.48\,\text{kVAR}$$

(Check: $S = \sqrt{P^2+Q^2} = \sqrt{8.83^2 + 5.48^2} \approx 10.39\,\text{kVA}$.)

(c) Two-wattmeter method (2 marks)

Two wattmeters are connected with their current coils in any two lines (say R and Y) and their voltage coils between that line and the third line (B). Total three-phase power $P = W_1 + W_2$ — valid for balanced or unbalanced loads (3-wire). The power factor is found from

$$\tan\phi = \sqrt{3}\,\frac{W_1 - W_2}{W_1 + W_2} \;\Rightarrow\; \cos\phi = \cos\!\left[\tan^{-1}\!\left(\sqrt{3}\frac{W_1-W_2}{W_1+W_2}\right)\right]$$

Series RLC with $R=10\,\Omega$, $L=0.1\,\text{H}$, $C=50\,\mu\text{F}$, supply $230\,\text{V}$, $50\,\text{Hz}$ ($\omega = 2\pi\times 50 = 314.16\,\text{rad/s}$).

Reactances:

$$X_L = \omega L = 314.16 \times 0.1 = 31.42\,\Omega$$ $$X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times 50\times 10^{-6}} = \frac{1}{0.01571} = 63.66\,\Omega$$

Net reactance $X = X_L - X_C = 31.42 - 63.66 = -32.24\,\Omega$ (capacitive).

(a) Impedance:

$$Z = \sqrt{R^2 + X^2} = \sqrt{10^2 + 32.24^2} = \sqrt{100 + 1039.4} = \sqrt{1139.4} = \boxed{33.76\,\Omega}$$

(b) Current and phase angle:

$$I = \frac{V}{Z} = \frac{230}{33.76} = \boxed{6.81\,\text{A}}$$ $$\phi = \tan^{-1}\!\frac{X}{R} = \tan^{-1}\!\frac{-32.24}{10} = -72.8^\circ$$

Since $X_C > X_L$, the current leads the voltage by $72.8^\circ$ (capacitive circuit).

(c) Resonant frequency:

$$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 50\times 10^{-6}}} = \frac{1}{2\pi\sqrt{5\times 10^{-6}}} = \frac{1}{2\pi \times 2.236\times 10^{-3}} = \boxed{71.18\,\text{Hz}}$$

(a) Average value, RMS value and form factor (3 marks)

Average value: The average (mean) value of an alternating quantity over one cycle (or half cycle for a symmetrical wave) — the steady value that transfers the same charge in the same time. For a sinusoid over a half-cycle:

$$I_{avg} = \frac{2 I_m}{\pi} = 0.637\,I_m$$

RMS (root-mean-square) value: The equivalent DC value that produces the same heating (power) in a resistance. For a sinusoid:

$$I_{rms} = \frac{I_m}{\sqrt{2}} = 0.707\,I_m$$

Form factor:

$$k_f = \frac{\text{RMS value}}{\text{average value}} = \frac{0.707\,I_m}{0.637\,I_m} = \frac{\pi}{2\sqrt{2}} = \boxed{1.11}$$

(b) Power factor (3 marks)

Power factor is the cosine of the phase angle $\phi$ between voltage and current, $\text{pf} = \cos\phi = \dfrac{P}{S} = \dfrac{\text{active power}}{\text{apparent power}}$. It indicates what fraction of the apparent power (VA) does useful work (W).

Why a low lagging pf is undesirable:

• For a given power $P$, line current $I = \dfrac{P}{V\cos\phi}$ increases as pf falls, so $I^2R$ line and transformer losses rise and efficiency drops.
• Larger current requires larger conductor size, switchgear and transformer kVA rating (higher cost).
• Causes greater voltage drop and poor voltage regulation.
• Utilities penalize consumers with low pf. (Remedied by power-factor correction using capacitors.)

(a) Electric–magnetic circuit analogy (3 marks)

(b) Reluctance and flux (3 marks)

Given: $l = 40\,\text{cm} = 0.4\,\text{m}$, $A = 5\,\text{cm}^2 = 5\times 10^{-4}\,\text{m}^2$, $\mu_r = 500$, $N = 200$, $I = 2\,\text{A}$, $\mu_0 = 4\pi\times 10^{-7}$.

Reluctance:

$$S = \frac{l}{\mu_0\mu_r A} = \frac{0.4}{(4\pi\times 10^{-7})(500)(5\times 10^{-4})}$$ $$= \frac{0.4}{3.1416\times 10^{-7}} = \boxed{1.273\times 10^{6}\,\text{AT/Wb}}$$

MMF: $\mathcal{F} = NI = 200\times 2 = 400\,\text{AT}$.

Flux:

$$\phi = \frac{\mathcal{F}}{S} = \frac{400}{1.273\times 10^6} = 3.14\times 10^{-4}\,\text{Wb} = \boxed{0.314\,\text{mWb}}$$

(a) EMF equation of a DC generator (3 marks)

Let $\phi$ = flux per pole (Wb), $P$ = number of poles, $N$ = speed (rpm), $Z$ = total armature conductors, $A$ = number of parallel paths ($A=P$ for lap, $A=2$ for wave).

Flux cut by one conductor in one revolution $= \phi P$. Time for one revolution $= \dfrac{60}{N}$ s. EMF per conductor:

$$e = \frac{\phi P N}{60}\,\text{V}$$

Conductors per parallel path $= Z/A$, so total generated EMF:

$$\boxed{E_g = \frac{\phi Z N P}{60 A}\,\text{V}}$$

(b) Calculation (3 marks)

Given: $P = 4$, lap-wound so $A = P = 4$, $Z = 440$, $N = 1000\,\text{rpm}$, $\phi = 25\,\text{mWb} = 25\times 10^{-3}\,\text{Wb}$.

$$E_g = \frac{\phi Z N P}{60 A} = \frac{(25\times 10^{-3})(440)(1000)(4)}{60\times 4}$$ $$= \frac{44000}{240} = \boxed{183.3\,\text{V}}$$

(a) Working principle and back EMF (3 marks)

Working principle: When a current-carrying conductor is placed in a magnetic field, it experiences a mechanical force $F = BIl$ (Lorentz force). In a DC motor, current from the supply flows through armature conductors situated in the field produced by the field poles; the resulting forces on all conductors produce a net torque that rotates the armature. The commutator reverses the current in each conductor as it passes the magnetic neutral axis, so the torque remains unidirectional.

Back EMF ($E_b$): As the armature rotates in the magnetic field, it also generates an EMF that, by Lenz's law, opposes the applied voltage — the back (counter) EMF, $E_b = \dfrac{\phi Z N P}{60 A}$. Its significance: it makes the motor self-regulating — armature current $I_a = \dfrac{V - E_b}{R_a}$ automatically adjusts to the load. At light load speed rises, $E_b$ rises, $I_a$ falls; at heavy load $E_b$ falls, $I_a$ rises to supply more torque.

(b) Calculation (3 marks)

Given: $V = 220\,\text{V}$, $I_a = 20\,\text{A}$, $R_a = 0.5\,\Omega$.

Back EMF:

$$E_b = V - I_a R_a = 220 - (20)(0.5) = 220 - 10 = \boxed{210\,\text{V}}$$

Mechanical power developed (gross):

$$P_m = E_b I_a = 210 \times 20 = 4200\,\text{W} = \boxed{4.2\,\text{kW}}$$

(a) PMMC instrument construction and working (3 marks)

Construction (diagram described): A lightweight rectangular moving coil is wound on an aluminium former and pivoted between the poles of a permanent magnet; a soft-iron cylindrical core is placed inside the coil to give a strong, uniform radial magnetic field. Two control springs (also carrying current to the coil) provide controlling torque and act as current leads, and a pointer moves over a scale.

Working: When current $I$ passes through the coil in the radial field $B$, each side experiences a force and a deflecting torque $T_d = N B I l b = G I$ (proportional to current) is produced. The coil rotates until $T_d$ equals the spring controlling torque $T_c = k\theta$, so $\theta \propto I$ — giving a uniform (linear) scale. Eddy-current damping in the aluminium former provides damping. It reads only DC (average value) because torque reverses with current.

(b) PMMC vs Moving-Iron (3 marks)

(PMMC is more accurate and consumes less power; MI is cheaper and robust.)

(a) Range extension (3 marks)

Ammeter (shunt): A moving-coil meter carries only a small full-scale current $I_m$. To read a large current $I$, a low-value shunt resistance $R_{sh}$ is connected in parallel with the coil so that most of the current bypasses the coil. Since coil and shunt have equal voltage:

$$I_m R_m = (I - I_m) R_{sh} \;\Rightarrow\; R_{sh} = \frac{I_m R_m}{I - I_m}$$

Multiplying power $n = I/I_m$, so $R_{sh} = \dfrac{R_m}{n-1}$.

Voltmeter (series multiplier): To measure a high voltage $V$, a high-value multiplier resistance $R_s$ is connected in series with the coil so that the coil sees only its rated drop $I_m R_m$:

$$V = I_m (R_m + R_s) \;\Rightarrow\; R_s = \frac{V}{I_m} - R_m = R_m(m-1),\quad m = \frac{V}{I_m R_m}$$

(b) Shunt calculation (3 marks)

Given: $R_m = 50\,\Omega$, $I_m = 1\,\text{mA} = 0.001\,\text{A}$, $I = 1\,\text{A}$.

$$R_{sh} = \frac{I_m R_m}{I - I_m} = \frac{0.001 \times 50}{1 - 0.001} = \frac{0.05}{0.999} = \boxed{0.05005\,\Omega \approx 0.05\,\Omega}$$

Superposition theorem (statement and limitations)

Statement: In a linear, bilateral network containing more than one independent source, the response (current or voltage) in any branch equals the algebraic sum of the responses produced by each independent source acting alone, with all other independent sources de-activated (voltage sources short-circuited, current sources open-circuited, internal resistances retained).

Limitations:

• Applies only to linear networks; cannot be used for non-linear elements (diodes, etc.).
• Cannot be used directly to compute power (power is proportional to $I^2$, not linear).
• Only independent sources are considered one at a time; dependent sources remain active throughout.

Current through the $5\,\Omega$ resistor

Circuit: $10\,\text{V}$ source in series with $2\,\Omega$, and a $4\,\text{A}$ source, both feeding the common $5\,\Omega$ branch (the $2\,\Omega$ and $5\,\Omega$ meeting at the load node, current source in parallel with the $5\,\Omega$).

Source 1 — 10 V acting alone (open the 4 A current source): Current flows through $2\,\Omega$ and $5\,\Omega$ in series:

$$I' = \frac{10}{2 + 5} = 1.429\,\text{A}$$

Source 2 — 4 A acting alone (short the 10 V source, leaving $2\,\Omega$): The $4\,\text{A}$ divides between $2\,\Omega$ and $5\,\Omega$ (current divider). Current through $5\,\Omega$:

$$I'' = 4 \times \frac{2}{2 + 5} = 4 \times \frac{2}{7} = 1.143\,\text{A}$$

Total (both adding through the $5\,\Omega$):

$$I_{5\Omega} = I' + I'' = 1.429 + 1.143 = \boxed{2.57\,\text{A}}$$

(If the current source were directed to oppose, the answer would be $1.429 - 1.143 = 0.286\,\text{A}$; the additive case giving 2.57 A is assumed here.)