Solution Electrical Numerical 19

Numerical 19: 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. supply 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.

Solution: Given: Resistance of solenoid (R) = 35 Ω

Voltage = 220 V

Frequency (f) = 50 Hz

Initial Current $\displaystyle \mathrm{I_1}$ = 2 A

Final Current $\displaystyle \mathrm{I_2}$ = 0.7 A

When Plunger out:

Impedance $\displaystyle \mathrm{(Z) = \frac{V}{I} = \frac{220}{2} = 110}$  Ω

and R = 35 Ω

$\displaystyle \mathrm{X = \sqrt{Z^2-R^2}\ =\ \sqrt{110^2-35^2}\ =\ 104.28 }$ Ω

But, $\displaystyle \mathrm{X =2\pi fL }$ 

$\displaystyle \mathrm{104.28 =2\pi 50L }$ 

∴ L = 0.332 H

When Plunger In:

Impedance $\displaystyle \mathrm{(Z) = \frac{V}{I} = \frac{220}{0.7} = 314.28}$  Ω

and R = 35 Ω

$\displaystyle \mathrm{X = \sqrt{Z^2-R^2}\ =\ \sqrt{314.28^2-35^2}\ =\ 312.3 }$ Ω

But, $\displaystyle \mathrm{X =2\pi fL }$ 

$\displaystyle \mathrm{312.3 =2\pi 50L }$ 

∴ L = 0.995 H

We know that

$\displaystyle \mathrm{L = \frac{N\phi }{I} }$ 

Here Nф = Flux linkage

thus, Flux linkage = L x I = 0.995 x √2 x 0.7 = 0.985 w-turn.