A 1.15 kilo Ohms resistor and a 590 mH inductor are connected in series to a 1350 Hz generator with an rms voltage of 12.1 V.
What is the rms current in the circuit in mA?
What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part A in nF?
The first part is correct; however, the second part was wrong
Mr. Un-couth
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XL = 2pi fl = (6.28)(1,350Hz)(.590H) = 5,002 Ohm
Z = 5,002 Ohm / sin (arc tan 5002/1150) = 5132 Ohms
Rms current in circuit = V/Z = 12.1V / 5132 Ohm = 2.36mA
new Z = (old Z) / 2 = 5132 Ohms / 2 = 2566 Ohm
XL- XC = [sin (arc cos R/new Z)] X (new Z) = [sin (arc cos 1150/2566)] X (2566 Ohm) = 2294 Ohm
XC = 5,002 Ohms - 2294 Ohms = 2708 Ohms
XC also = 1 / 2pi fC
C = 1 / (6.28)(1350Hz)(2708 Ohm) = .000000044F = 44nF
Response to additional details:
I got a little carried away I guess. I took half the old Z instead of halving the current. Maybe this will yield the correct solution. Sorry for the inconvenience.
XL - XC = [sin (arc cos R/new Z)] X (new Z) = [sin (arc cos 1150/10,264)] x 10,264 Ohms = 10,199 Ohms. This is greater than XL which means it is in the 4th quad and the total reactance is now capacitive. So the added capacitor must have a capacitance value such that it`s reactance can cancel 5,002 Ohms of XL and add 10,199 Ohms of capacitive reactance. Therefore the absolute value of XC = XL - (- XC) = 5,002 Ohms + 10,199 Ohms = 15,201 Ohms
15,201 Ohms = 1 /(6.28) (1350 Hz) (C)
C = 1 / (6.28) (1350) (15,201) = 8nF