Question
An inductive coil with a resistance R and Inductance L is connected in parallel with a 30ohm resistor. The combination is then connected across a 60hz ac source. If the currents in the coil, resistor and the total are 6,4 and 8A respectively. Determine the inductance L of the coil?
Answers
bobpursley
Parallel. assume some V across the parallel. but Ir=6 amps, so V must be 180V (R=30ohms)
then Xl=180/4=45 ohms
and Z=180/8=22.5 ohms
but Xl=2PI*f*L or L=45/2*PI*60 Henrys
then Xl=180/4=45 ohms
and Z=180/8=22.5 ohms
but Xl=2PI*f*L or L=45/2*PI*60 Henrys
What is the PI
bobpursley
PI? 3.141592....
Surely you have done impedance problems before...
Surely you have done impedance problems before...
Given:
R1 = 30 Ohms.
I1 = 4A.
Ic = 6A = Current through coil.
I = 8A = Total current.
E = I1*R1 = 4 * 30 = 120 Volts. = Applied voltage = Voltage across R1.
WL = E/Ic = 120/6 = 20 Ohms. = Inductive reactance.
L = 20/W = 20/377 = 0.053 henrys.
Note: W = 2pi*F = 6.28 * 60 = 377..
R1 = 30 Ohms.
I1 = 4A.
Ic = 6A = Current through coil.
I = 8A = Total current.
E = I1*R1 = 4 * 30 = 120 Volts. = Applied voltage = Voltage across R1.
WL = E/Ic = 120/6 = 20 Ohms. = Inductive reactance.
L = 20/W = 20/377 = 0.053 henrys.
Note: W = 2pi*F = 6.28 * 60 = 377..
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