BE Computer Engineering (Pokhara University) Basic Electrical Engineering (PU, ELE 120) Question Paper 2079 Nepal

(a) Mesh (Loop) Analysis (7 marks)

Circuit: The 24 V source feeds through 4 Ω into the common node; the 12 V source feeds through 6 Ω into the same node; the 8 Ω resistor connects that node to ground.

Let the common-node-to-ground voltage be $V$. Using two loops with mesh currents $I_1$ (left loop, through 24 V and 4 Ω) and $I_2$ (right loop, through 12 V and 6 Ω), both returning through the 8 Ω resistor:

Loop 1:

$$24 = 4I_1 + 8(I_1 + I_2)$$ $$12I_1 + 8I_2 = 24 \quad (1)$$

Loop 2:

$$12 = 6I_2 + 8(I_1 + I_2)$$ $$8I_1 + 14I_2 = 12 \quad (2)$$

Solving (1) and (2):

From (1): $I_1 = \dfrac{24 - 8I_2}{12} = 2 - 0.667I_2$.

Substitute into (2):

$$8(2 - 0.667I_2) + 14I_2 = 12$$ $$16 - 5.33I_2 + 14I_2 = 12 \Rightarrow 8.667I_2 = -4 \Rightarrow I_2 = -0.461\,\text{A}$$ $$I_1 = 2 - 0.667(-0.461) = 2.308\,\text{A}$$

Currents supplied:

• Current from 24 V source: $I_1 = 2.31\,\text{A}$
• Current from 12 V source: $I_2 = -0.46\,\text{A}$ (i.e. 0.46 A flows into the 12 V source — it is being charged).

(b) Verification by Nodal Analysis and Power (7 marks)

Take the common node voltage $V$ (ground = 0). KCL at the node (sum of currents leaving = 0):

$$\frac{V-24}{4} + \frac{V-12}{6} + \frac{V}{8} = 0$$

Multiply by 24 (LCM):

$$6(V-24) + 4(V-12) + 3V = 0$$ $$6V - 144 + 4V - 48 + 3V = 0 \Rightarrow 13V = 192 \Rightarrow V = 14.77\,\text{V}$$

Current through 8 Ω resistor (node to ground):

$$I_{8} = \frac{V}{8} = \frac{14.77}{8} = 1.846\,\text{A}$$

Cross-check with mesh result: $I_1 + I_2 = 2.308 + (-0.461) = 1.847\,\text{A}$ ✓ (agreement).

Power dissipated in 8 Ω:

$$P_8 = I_8^2 R = (1.846)^2 \times 8 = 27.3\,\text{W}$$

(Equivalently $P_8 = V^2/R = 14.77^2/8 = 27.3$ W.)

(a) Thevenin's Theorem (4 marks)

Statement: Any linear, bilateral two-terminal network containing sources and resistances can be replaced, as seen from its two output terminals, 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 looking back into the network with all independent sources deactivated.

Procedure:

1. Remove the load from terminals A-B.
2. Find the open-circuit voltage across A-B — this is $V_{Th}$.
3. Deactivate all independent sources (short voltage sources, open current sources), keeping their internal resistances.
4. Compute the equivalent resistance looking into A-B — this is $R_{Th}$.
5. Redraw as $V_{Th}$ in series with $R_{Th}$ feeding the load.

(b) Thevenin Equivalent (5 marks)

Circuit: 20 V source in series with 5 Ω, with 10 Ω across A-B.

$V_{Th}$ (open-circuit voltage across A-B = voltage across 10 Ω): With load removed, the 10 Ω is across the terminals. Current in the loop $= 20/(5+10) = 1.333$ A.

$$V_{Th} = 1.333 \times 10 = 13.33\,\text{V}$$

$R_{Th}$: Short the 20 V source. Then 5 Ω and 10 Ω appear in parallel across A-B:

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

Thevenin equivalent: $V_{Th} = 13.33\,\text{V}$ in series with $R_{Th} = 3.33\,\Omega$.

(c) Maximum Power Transfer (3 marks)

Maximum power is transferred when the load resistance equals the Thevenin resistance:

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

Maximum power:

$$P_{max} = \frac{V_{Th}^2}{4R_{Th}} = \frac{13.33^2}{4 \times 3.33} = \frac{177.8}{13.33} = 13.33\,\text{W}$$

(a) Star vs Delta Connection (6 marks)

Star derivation (voltage): Line voltage is the phasor difference of two phase voltages 120° apart:

$$V_L = 2V_{ph}\cos 30^\circ = \sqrt{3}\,V_{ph}, \qquad I_L = I_{ph}$$

Delta derivation (current): Line current is the phasor difference of two phase currents 120° apart:

$$I_L = 2I_{ph}\cos 30^\circ = \sqrt{3}\,I_{ph}, \qquad V_L = V_{ph}$$

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

(b) Balanced Star Load Calculation (8 marks)

Given: $Z_{ph} = 8 + j6\,\Omega$, $V_L = 400$ V, star-connected.

Phase impedance magnitude and angle:

$$|Z| = \sqrt{8^2 + 6^2} = 10\,\Omega, \qquad \phi = \tan^{-1}(6/8) = 36.87^\circ$$

Phase voltage (star):

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

Line current ( = phase current in star):

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

Power factor:

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

Total active power:

$$P = \sqrt{3}\,V_L I_L \cos\phi = \sqrt 3 \times 400 \times 23.09 \times 0.8 = 12.8\,\text{kW}$$

Total reactive power:

$$Q = \sqrt{3}\,V_L I_L \sin\phi = \sqrt 3 \times 400 \times 23.09 \times 0.6 = 9.6\,\text{kVAR}$$

Summary: $I_L = 23.1$ A, p.f. $= 0.8$ lag, $P = 12.8$ kW, $Q = 9.6$ kVAR.

(a) EMF Equation & Principle (5 marks)

Principle (mutual induction): A transformer has two coils (primary and secondary) wound on a common magnetic core. When alternating voltage is applied to the primary, it drives an alternating flux in the core. This time-varying flux links the secondary winding and, by Faraday's law, induces an EMF in it. Energy is transferred from primary to secondary purely by mutual magnetic coupling, with no electrical connection.

EMF equation: Let the flux be $\phi = \phi_m \sin\omega t$. The EMF induced per turn is $-d\phi/dt$. For $N$ turns the peak EMF is $N\omega\phi_m = N(2\pi f)\phi_m$. The RMS value is

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

Thus $E_1 = 4.44\,f\,N_1\,\phi_m$ and $E_2 = 4.44\,f\,N_2\,\phi_m$, giving the turns ratio $\dfrac{E_1}{E_2} = \dfrac{N_1}{N_2}$.

(b) Numerical (4 marks)

Given: 50 kVA, 2200/220 V, 50 Hz, $N_1 = 400$.

Secondary turns:

$$N_2 = N_1 \frac{V_2}{V_1} = 400 \times \frac{220}{2200} = 40\ \text{turns}$$

Maximum flux (from $E_1 = 4.44 f N_1 \phi_m$):

$$\phi_m = \frac{V_1}{4.44 f N_1} = \frac{2200}{4.44 \times 50 \times 400} = 0.0248\,\text{Wb} = 24.8\,\text{mWb}$$

Full-load currents:

$$I_1 = \frac{kVA}{V_1} = \frac{50000}{2200} = 22.7\,\text{A}, \qquad I_2 = \frac{50000}{220} = 227.3\,\text{A}$$

(c) Approximate Equivalent Circuit Referred to Primary (3 marks)

Description (referred to primary):

  o---R01---X01---+-----------o
  |               |
V1            Rm | Xm (shunt)   V2'  (= a*V2)
  |               |
  o---------------+-----------o

• $R_{01} = R_1 + a^2 R_2$ : total resistance referred to primary
• $X_{01} = X_1 + a^2 X_2$ : total leakage reactance referred to primary
• $R_m$ : core-loss (magnetising) resistance branch
• $X_m$ : magnetising reactance branch
• $a = N_1/N_2$ : turns ratio; the shunt (no-load) branch is moved to the input terminals for the approximate circuit.

Series RLC Circuit (6 marks)

Given: $R = 10\,\Omega$, $L = 0.1$ H, $C = 100\,\mu\text{F}$, $V = 230$ V, $f = 50$ Hz, $\omega = 2\pi f = 314.16$ rad/s.

Inductive reactance:

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

Capacitive reactance:

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

Net reactance: $X = X_L - X_C = 31.42 - 31.83 = -0.41\,\Omega$.

Impedance:

$$Z = \sqrt{R^2 + X^2} = \sqrt{10^2 + (-0.41)^2} = 10.008 \approx 10.0\,\Omega$$

Current:

$$I = \frac{V}{Z} = \frac{230}{10.0} = 22.98\,\text{A} \approx 23\,\text{A}$$

Power factor:

$$\cos\phi = \frac{R}{Z} = \frac{10}{10.0} = 0.999 \approx 1$$

Nature: Since $X_C > X_L$ (net reactance negative), the circuit is slightly capacitive (current leads voltage). It is operating very close to resonance ($f_0 = 1/(2\pi\sqrt{LC}) = 50.3$ Hz), so the power factor is nearly unity.

Definitions (and numerical) (6 marks)

RMS value: The RMS (root-mean-square) or effective value of an alternating quantity is the equivalent DC value that produces the same heating effect in a given resistance. For a sinusoid, $V_{rms} = V_m/\sqrt2 = 0.707\,V_m$.

Average value: The average (mean) value over a half cycle of a symmetrical alternating quantity. For a sinusoid, $V_{avg} = 2V_m/\pi = 0.637\,V_m$.

Form factor: The ratio of RMS value to average value: $K_f = V_{rms}/V_{avg}$. For a sinusoid, $K_f = 0.707/0.637 = 1.11$.

For $v(t) = 311\sin(314t)$

• Peak value: $V_m = 311\,\text{V}$
• RMS value: $V_{rms} = 311/\sqrt2 = 219.9 \approx 220\,\text{V}$
• Angular frequency / frequency: $\omega = 314$ rad/s, so $f = \omega/2\pi = 314/6.283 = 50\,\text{Hz}$
• Form factor: $K_f = 1.11$ (for a sine wave)

This is the standard 220 V, 50 Hz domestic supply.

Magnetic Circuit of an Iron Ring (6 marks)

Given: mean length $l = 40\,\text{cm} = 0.4$ m, area $A = 5\,\text{cm}^2 = 5\times10^{-4}\,\text{m}^2$, $\mu_r = 1000$, $N = 200$, $I = 2$ A. ($\mu_0 = 4\pi\times10^{-7}$).

MMF (magnetomotive force):

$$\mathcal{F} = NI = 200 \times 2 = 400\,\text{AT (ampere-turns)}$$

Reluctance:

$$S = \frac{l}{\mu_0 \mu_r A} = \frac{0.4}{(4\pi\times10^{-7})(1000)(5\times10^{-4})}$$ $$S = \frac{0.4}{6.283\times10^{-7}} = 6.366\times10^{5}\,\text{AT/Wb}$$

Flux:

$$\phi = \frac{\mathcal{F}}{S} = \frac{400}{6.366\times10^{5}} = 6.28\times10^{-4}\,\text{Wb} = 0.628\,\text{mWb}$$

Summary: MMF $= 400$ AT, $S = 6.37\times10^5$ AT/Wb, $\phi \approx 0.628$ mWb.

EMF Equation of a DC Generator (6 marks)

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

In one revolution each conductor cuts flux $= \phi P$ webers. Time for one revolution $= 60/N$ seconds. Average EMF per conductor:

$$e = \frac{\phi P}{60/N} = \frac{\phi P N}{60}$$

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

$$\boxed{E = \frac{\phi Z N P}{60 A}}$$

For lap winding $A = P$; for wave winding $A = 2$.

Numerical

Given: $P = 4$, lap-wound so $A = P = 4$, $Z = 600$, $N = 1000$ rpm, $\phi = 20\,\text{mWb} = 0.02$ Wb.

$$E = \frac{\phi Z N P}{60 A} = \frac{0.02 \times 600 \times 1000 \times 4}{60 \times 4}$$ $$E = \frac{48000}{240} = 200\,\text{V}$$

Generated EMF $= 200$ V.

(a) Principle & Back EMF (3 marks)

Principle: When a current-carrying conductor is placed in a magnetic field, it experiences a mechanical force $F = BIl$ (Lorentz/motor principle). In a DC motor, current-carrying armature conductors lie in the field of the poles; the resulting force on each conductor produces a torque that rotates the armature. The commutator reverses the current in each conductor as it passes the magnetic neutral so that the torque always acts in the same direction.

Back EMF ($E_b$): As the armature rotates in the magnetic field it cuts flux and generates an EMF that, by Lenz's law, opposes the applied voltage — hence "back" EMF, $E_b = \phi Z N P/60A$. Its significance: it limits and regulates the armature current ($I_a = (V - E_b)/R_a$) and makes the motor self-regulating — when load increases, speed and $E_b$ fall, allowing more current and torque to be drawn automatically.

(b) Numerical (3 marks)

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

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

Back EMF $= 210$ V.

PMMC Instrument (6 marks)

Construction (sketch in words): A rectangular coil of fine insulated wire is wound on a light aluminium former and pivoted on jewelled bearings between the poles of a permanent magnet. A soft-iron cylindrical core is placed inside the coil to make the air-gap field radial and uniform. Two spiral hair-springs (top and bottom) provide the controlling torque and also carry current to the coil. A pointer attached to the coil moves over a calibrated scale.

Working principle: When current $I$ flows through the coil placed in the radial magnetic field $B$, each side experiences a force, producing a deflecting torque

$$T_d = N B I L b = G I$$

where $G = NBLb$ is a constant. The hair-springs provide a controlling torque $T_c = K\theta$. At balance $T_d = T_c$, giving

$$\theta = \frac{GI}{K} \propto I$$

so the deflection is directly proportional to the current and the scale is uniform (linear). Damping is provided by eddy currents in the aluminium former.

Advantages (any two):

1. Uniform (linear) scale.
2. High sensitivity and accuracy; low power consumption.
3. No hysteresis error; effective eddy-current damping.

Disadvantages (any two):

1. Can measure only DC (gives zero average deflection for AC unless used with a rectifier).
2. More expensive due to the permanent magnet and precise construction; ageing of magnet/springs can cause errors.

Wheatstone Bridge (6 marks)

Circuit (described): Four resistors form a diamond/bridge: $P$ and $Q$ are the ratio arms, $R$ is a known standard variable resistance, and $S$ is the unknown resistance. A DC supply (battery) is connected across one diagonal (A-C) and a sensitive galvanometer $G$ across the other diagonal (B-D).

        A
      P/ \Q
      B---G---D       (galvanometer between B and D)
      R\ /S
        C
   (battery across A-C)

Working: The variable arm $R$ is adjusted until the galvanometer shows zero deflection. At this balanced condition no current flows through the galvanometer, so points B and D are at the same potential.

Balance condition (derivation): At balance, $I_g = 0$, so the same current $I_1$ flows through $P$ and $R$, and $I_2$ through $Q$ and $S$.

Potential at B equals potential at D:

$$I_1 P = I_2 Q \quad (1)$$ $$I_1 R = I_2 S \quad (2)$$

Dividing (1) by (2):

$$\frac{P}{R} = \frac{Q}{S} \Rightarrow \boxed{\dfrac{P}{Q} = \dfrac{R}{S}}$$

Hence the unknown resistance:

$$S = \frac{Q}{P}\,R$$

The bridge is a null (deflection-independent) method, so its accuracy does not depend on the galvanometer calibration or supply voltage.

Superposition Theorem (6 marks)

Statement: In a linear bilateral network containing more than one independent source, 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 (voltage sources short-circuited, current sources open-circuited).

Worked Example

Consider the 5 Ω resistor (call it $R$) fed by a 10 V source through a series 10 Ω resistor, with a 2 A current source connected across the 5 Ω resistor (a representative arrangement consistent with the data).

Source 1 — 10 V source acting alone (open the 2 A current source): Current flows through 10 Ω in series with 5 Ω:

$$I' = \frac{10}{10 + 5} = 0.667\,\text{A (through 5 Ω)}$$

Source 2 — 2 A current source acting alone (short the 10 V source): The 2 A splits between the 10 Ω (now to ground) and the 5 Ω in parallel. Current through the 5 Ω by current divider:

$$I'' = 2 \times \frac{10}{10 + 5} = 1.333\,\text{A}$$

Total current through 5 Ω (both contributions in the same direction):

$$I = I' + I'' = 0.667 + 1.333 = 2.0\,\text{A}$$

Note: If the two contributions oppose, take the algebraic difference. The method illustrated — solving for each source separately and superposing — is the key technique; with this configuration $I_{5\Omega} = 2\,\text{A}$.

Two-Wattmeter Method (4 marks)

Method: Two wattmeters are used to measure total power in a three-phase, three-wire system (balanced or unbalanced). Each wattmeter's current coil is connected in one line (say R and Y) and its pressure (voltage) coil is connected between that line and the third line (B). The total three-phase power is the algebraic sum of the two readings:

$$P = W_1 + W_2$$

For a balanced load with power factor angle $\phi$:

$$W_1 = V_L I_L \cos(30^\circ - \phi), \qquad W_2 = V_L I_L \cos(30^\circ + \phi)$$

so $W_1 + W_2 = \sqrt{3}\,V_L I_L \cos\phi$ (total power), which is correct for any power factor.

Power factor from the readings: Subtracting,

$$W_1 - W_2 = V_L I_L \sin\phi$$ $$\tan\phi = \sqrt{3}\,\frac{W_1 - W_2}{W_1 + W_2}$$

Hence the load power factor is

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

If $W_1 = W_2$, p.f. $= 1$ (unity). If one wattmeter reads zero, p.f. $= 0.5$; if one reads negative (p.f. below 0.5), its reading is subtracted.