45.MET April 2019 Q.8(b)

The low-voltage release of an a.c. motor-starter consists of a solenoid into which an iron plunger is drawn against a spring. The resistance of the solenoid is 35 ohm. When connected to a 220 V, 50 Hz, a.c. 2supply the current taken is at first 2A, and when the plunger is drawn into the “full-in” position the current falls to 0.7 A. Calculate the inductance of the solenoid for both positions of the plunger, and the maximum value of flux-linkages in weber-turns for the “full-in” position of the plunger.

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Given

R = 35 ohm

V = 220 V

F = 50 Hz

Iinitial = 2 A

Ifinal = 0.7 A

To find

1. Inductance of solenoid at initial position

2. Inductance of solenoid at final(plunger in) position

3. Maximum value of flux linkage in weber turns

Solution

Initial Impedence Zinitial  = V/ Iinitial

                                         = 220 / 2

                                         = 110 ohm

Impedence 2 = Reactance2 + Resistance2

Reactance Xinitial  = √ (Zinitial2 – Resistance2)

                               = √ (1102 - 352)

                               = 104.25 ohm

            X = 2 x Pi x F x L

   104.25 = 2 x 3.14 x 50 x Linitial

1. Linitial = 104.225 / (2 x 3.14 x 50)

               = 0.332 Henry

When plunger is in Impedence Zfinal  =  V / Ifinal

                                                             = 220 / 0.7

                                                             = 314.28 ohm

Reactance Xfinal = √ (Zfinal2 – Resistance
                            = √ (314.282 - 352)

                            = 312.3 ohm

          X = 2 x Pi x F x L

   312.3  = 2 x 3.14 x 50 x Lfinal

2. Lfinal = 312.3 / (2 x 3.14 x 50)

              = 0.995 Henry

3. Maximum flux linkage = inductance x peak value of current

                                       = 0.995 x √ 2 x 0.7

                                       =  0.985 Weber Turns