### Important note

*There is something wrong with this question paper. It seems that instead of "Design of Electrical Systems" the IEI has given problems from "Electrical Machines". You should raise a complaint to director_eea@ieindia.org in this regard.*** **

**Numerical**

**A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb? (AMIE Summer 2023, 8 marks)**

**Solution**

The back EMF (E_{b}) of a DC motor can be calculated using the formula:

E_{b} = V - I_{a}R_{a}

Given:

V = 120 V

I_{a} = 200 A

R_{a} = 0.02 ohms

Substituting the values into the formula:

E_{b} = 120 − 200 × 0.02 = 120 − 4 = 116 V

Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ_{p}).

The formula to calculate the speed of a DC motor is:

N = 60×E_{b}/(P×φ_{p})

Where P is the number of poles

= 2 (assuming it's a 2-pole motor)

φ_{p} = 0.04 Wb

Substituting the values into the formula:

N = 60×116/2×0.04 = 87000 RPM

**Numerical**

**A 50 kW, 250 V shunt generator operates on full load at 1500 rpm. The armature has 6 poles and is lap wound with 200 turns. Find the induced emf and the flux/pole at full load. Given that the armature and field resistances are 0.01 and 125 Ω respectively. Neglect armature reaction. (AMIE Summer 2023, 8 marks)**

**Solution**

I_{L} = (50 x 10^{3})/250 = 200 A

I_{sh} = V/R_{sh} = 250/125 = 2 A

I_{a} = I_{L} + I_{sh} = 202 A

Induced emf, E_{g} = V + I_{a}R_{a}

= 250 + 202 x 0.01 = 252.02 V

Z = 200 x 2 (As one turn = 2 conductors)

N = 1500 rpm

P = A = 6 (lap wound)

φ = E_{g}(60)A/PZN

= 252.02 x 60 x 6/(400 x 1500 x 6)

= 0.025205 Wb

= 25.205 mWb

**Numerical**

**A 250 V, 25 kW, 4-pole DC generator has 328 wave connected armature conductors. When the machine is delivering full load, the brushes are given a lead of 7.2 electrical degrees. Calculate the cross magnetizing amp-turns/pole. (AMIE Summer 2023, 7 marks)**

**Solution**

Given data:

Terminal Voltage V = 250 V

Output Power Pout = 25 kW

No. of Poles P = 4

No. of conductors Z = 328

θ_{e} = 7.2°

Wave connected A = 2

Cross-magnetising ampere-turns/pole. = ?

Load current supplied

I_{a} = (25 × 1000) / 250 = 100A

I = 100 / A = 100 / 2 = 50A (current/path)

\({\theta _m} = \frac{{2{\theta _e}}}{P} = \frac{{2x7.2}}{4} = {3.6^0}\)

Cross-magnetizing ampere-tums/pole

\(A{T_c}/Pole = ZI\left( {\frac{1}{{2p}} - \frac{{{\theta _m}}}{{360}}} \right)\)

\( = 328x50\left( {\frac{1}{{2x4}} - \frac{{3.6}}{{360}}} \right) = 1886\)

**Numerical**

**A 220 V DC Shunt Motor runs at 500 rpm when the armature current is 50 A. Calculate the speed, if torque is doubled. Given the R = 0.2 ohm.(AMIE Summer 2023, 7 marks)**

**Solution**

Performance eqn of DC shunt motors are

E_{b} = V_{t} - I_{a}R_{a}

or, E_{b} = K_{a}φω

and τ = KaφI_{a}

E_{b1} = 220 - 50 × 0.2 = 210 V

Since the torque is double i.e., the armature current is doubled

i.e., I_{a2} = 2 × I_{a1} = 2 × 50 = 100 A

∴ E_{b2} = 220 - 100 × 0.2 = 200 V

E_{b2}/E_{b1} = N_{2}/N_{1}

⇒N_{2} = N_{1} x (E_{b2}/E_{b1})

= 500 x (200/210)

= 476 rpm

**Numerical**

**A single-phase transformer has 600 primary and 80 secondary turns. The mean length of the flux path in the ferromagnetic core is 1.6 m, the value of flux in the core for a magnetic field strength of 1.2 T is 425 AT/m, and the corresponding core loss is 1.5 W/kg at 50 Hz. The density of the core is 7400 kg/m3. If the maximum value of flux density is 1.2 T when the primary is connected to a 3300-V, 50-Hz supply, calculate (a) the cross-sectional area of the core, (b) the secondary voltage on no load, (c) the primary magnetizing current, and (d) the core loss. (AMIE Summer 2023, 8 marks)**

**Solution**

(a) 3300=4.44 × 600 × 50 × Ø_{m}

Ø_{m} = 0.0248 Wb

Cross-sectional area of the core

= Ø_{m}/B_{m} = 0.0248/1.2

= 0.02067 m^{2}

(b) Secondary voltage on load = 3300 × (80/600)= 440 V

(c) Primary magnetizing current = mmf/ primary turns

= (H × l)/600

= (425 × 1.6 )/ 600

= 1.133 A

Assuming sinusoidal current, the RMS value of the magnetizing current is

= 1.133/√2= 0.80 A

(d) Volume of the core = 1.6 × 0.02067 = 0.03307 m^{3}

Mass of the core= 0.03307 × 7400 = 244.73 kg

Therefore, the core loss = 244.73 × 1.5= 367 W

**Numerical**

**A 2400V/400V single-phase transformer takes a no-load current of 0.5 A and the core losses are 400 W. Determine the values of magnetising and core loss components of the no-load current. (AMIE Summer 2023, 7 marks)**

**Solution**

V_{1} = 2400 V

V_{2} = 400 V

I_{0} = 0.5 A

Core loss (i.e. iron loss)

= 400 = V_{1}I_{0}cosφ_{0}

Hence, 400 = 2400 x 0.5 x cosφ_{0}

From this, cosφ_{0} = 0.3333

Hence, φ_{0} = 70.53^{0}

The no-load diagram is shown below.

Magnetising component

I_{M} = I_{0}sinφ_{0}

= 0.5sin(70.53^{0})

= 0.471 A

Core loss component

I_{C} = I_{0}cosφ_{0}

= 0.5cos(70.53^{0})

= 0.167 A

**Numerical**

**A 440 V, 3-phase, 50 Hz, 4-pole, Y-connected induction motor has a full-load speed of 1425 r.p.m. The rotor has an impedance of (0 4 + j 4) ohm per phase and rotor/stator turn ratio is 0.8. Calculate (i) full-load torque (ii) rotor current and (Hi) full-load rotor Cu loss. (AMIE Summer 2023)**

**Solution**

N_{s} = 120f/P = 120 x 50/4

1500 rpm = 25 rps

s = (N_{s} - N)/N_{s}

= (1500 - 1425)/1500

= 0.05

E_{1} = 440√3 = 254 V/phase

E_{2} = KE_{1} = 0.8 x 254 = 203.2 V

Full load torque

\({T_f} = \frac{3}{{2\pi {N_s}}}x\frac{{sE_s^2{R_2}}}{{R_2^2 + {{(s{X_2})}^2}}}\)

Putting values,

T_{f }= 78.87 N-m

Rotor current

\(I_2^ = \frac{{s{E_2}}}{{\sqrt {{{({R_2})}^2} + {{(s{X_2})}^2}} }} = \frac{{0.05x203.2}}{{\sqrt {{{0.4}^2} + {{(0.05x4)}^2}} }}\)

= 22.73 A

^{Total copper loss}

_{2}’)

^{2}R

_{2}= 3 x 22.73

^{2}x 4

**AMIE/BTech/Junior Engineer exams**is available at

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