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rjedlicka rjedlicka
wrote...
Posts: 97
Rep: 1 0
11 years ago
An inductance coil draws 2.8 amps of dc current when connected to a 45V battery. When connected to a voltage source with waveform V=120cos120t, the maximum current drawn is 4.6amps. Calculate the resistance and inductance of the coil.
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wrote...
11 years ago
R = 45/2.8 = 16 ohms.

Inductive impedance = 2pi f L
                              = 120/4.6 = 2.61 ....as it was peak reading meter

now, 120t = 2pift => 2pif = 120

L= 2.61/120 = 0.02175 henris = 21.75 mH
wrote...
11 years ago
At the instant when the coil is first connected to the battery;
 maximum current flows and its magnitude is only limited by
 the coils intrinsic resistance.

 By Ohm's law:  R = V/i = 45/2.8 = 16.07 ?.

 When the coil is connected to the ac voltage source; the coils
 inductive reactance determines the maximum current drawn
 from the source.

 By Ohm's law:  X = V/i

 where, X is the coil's inductive reactance  and is given by:

 X = 2?fL = V/i

where,
f is the frequency: f = 120/2? = 60/?
L is the coil's inductance.

L = 120 / [(2?60/?)4.6] = 1/4.6 = 0.2174 H
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