**Design the 5 sections of a 6-stud starter for a 3-phase slip-ring induction motor The full-load slip is 2% and the maximum starting current is limited to twice the full-load current. Rotor resistance per phase is 0.03 Ω. (AMIE Winter 2023)**

Solution

Full load slip = 2%

Slip at 2 times full load current = 2 x 2 = 4% = 0.04

\(\alpha = {(0.04)^{1/n}} = {0.04^{1/5}} = 0.76\)

\({R_1} = \frac{{0.02}}{{0.04}} = 0.5\)

Resistances of the sections are

\(\begin{array}{l}{r_1} = (1 - \alpha ){R_1} = (1 - 0.76)(0.5) = 0.12\\{r_2} = \alpha {r_1} = 0.76x0.12 = 0.0912\\{r_3} = {\alpha ^2}{r_1} = {(0.76)^2}(0.12) = 0.693\\{r_4} = {\alpha ^3}{r_1}\\{r_5} = {\alpha ^4}{r_1}\end{array}\)

**In** **a transformer, the core loss is found to be 52 W at 40 Hz and 90 W at 60 Hz measured at same peak flux density. Compute the hysteresis and eddy current losses at 50 Hz. (AMIE Winter 2023).**

Solution

\(\begin{array}{l}{W_i} = Af + B{f^2}\\\frac{{{W_i}}}{f} = A + Bf\\\frac{{52}}{{40}} = A + 40B\\and\\\frac{{90}}{{60}} = A + 60B\\\therefore \,A = 0.9;B = 0.01\end{array}\)

At 50 Hz, two losses are

\(\begin{array}{l}{W_h} = {A_f} = 0.9x50 = 45\,W\\{W_e} = B{f^2} = 0.01x{50^2} = 25\,W\end{array}\)

**A** **short-circuit test, when performed on the h.v. side of a 10 kVA, 2000/400 V single phase transformer, gave the following data ; **

**60 V, 4 A, 100 IV**

**If the l.v. side is delivering full load current at 0.8 p.f lag and at 400 V, find the voltage applied to h.v. side. (AMIE Winter 2023)**

Solution

\(\begin{array}{l}{Z_{01}} = 60/4 = 15\Omega \\{R_{01}} = 100/{4^2} = 6.25\Omega \\{X_{01}} = \sqrt {{{15}^2} - {{6.25}^2}} = 13.63\Omega \\FL,\,{I_1} = 10,000/2000 = 5A\end{array}\)

Total transformer voltage drop as referred to primary is

\({I_1}({R_{01}}\cos \phi + {X_{01}}\sin \phi ) = 5(6.25\,x\,0.8 + 13.63\,x\,0.6) = 67V\)

Hence, primary voltage has to be raised from 2000 V to 2067 V in order to compensate for the total voltage drop in the transformer. In that case secondary voltage on load would remain the same as on no-load.

**A commutator with a diameter of 50 cm rotates at 1000 rpm. For a brush width of 1.5 cm. find the time of commutation. (AMIE Winter 2023)**

Solution

Peripheral velocity of commutator is

\(V = \pi DN = \pi (50)\left( {\frac{{1000}}{{60}}} \right)\,cm/s\)

Now, brush width = v x time of commutation

Hence, time of commutation = brush width/v

= \(\frac{{(1.5/100)}}{{\pi (50)(1000/60)}} = 0.573\,ms\)

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