J
jalbers@bsu.edu
Guest
I am trying to understand XLR circuits and the use of complex numbers
to find solutions. I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing...
Suppose that a 50 ohm resistor, 1 mH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.
Z = 50 + 62.8j 159j
Z = 50 96.2j or 108.417@-62.53
Suppose that I wanted to find the voltage drops across L, R, and C
when the power source is at .707V or 1@45 or .707 + .707j .
I = V/Z or
I = 1@45 / 108.417@-62.53 = .0092@107.53 or -.0027+.0087j
Taking the real part of -.0027+.0087j means that when the voltage
source is at .707 volts, the current through L, R, and C is at -.0027
amps.
The voltage across R:
(50@0) * (.0092@107.53) = .46@107.53 or -.138 + .438j
Taking the real part of -.138 + .438j means that when the voltage
source is at .707 volts, the voltage across R is at -.138 volts.
The voltage across L:
(62.8@90) * (.0092@107.53) = .577@197.53 or -.550 - .173j
Taking the real part of -.550 - .173j means that when the voltage
source is at .707 volts, the voltage across L is at -.550 volts.
The voltage across C:
(159@-90) * (.0092@107.53) = 1.4628@17.53 or 1.394 + .440j
Taking the real part of 1.394 + .440j means that when the voltage
source is at .707 volts, the voltage across L is at 1.394 volts.
------
Any comments or corrections would be greatly appreciated. Thanks
to find solutions. I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing...
Suppose that a 50 ohm resistor, 1 mH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.
Z = 50 + 62.8j 159j
Z = 50 96.2j or 108.417@-62.53
Suppose that I wanted to find the voltage drops across L, R, and C
when the power source is at .707V or 1@45 or .707 + .707j .
I = V/Z or
I = 1@45 / 108.417@-62.53 = .0092@107.53 or -.0027+.0087j
Taking the real part of -.0027+.0087j means that when the voltage
source is at .707 volts, the current through L, R, and C is at -.0027
amps.
The voltage across R:
(50@0) * (.0092@107.53) = .46@107.53 or -.138 + .438j
Taking the real part of -.138 + .438j means that when the voltage
source is at .707 volts, the voltage across R is at -.138 volts.
The voltage across L:
(62.8@90) * (.0092@107.53) = .577@197.53 or -.550 - .173j
Taking the real part of -.550 - .173j means that when the voltage
source is at .707 volts, the voltage across L is at -.550 volts.
The voltage across C:
(159@-90) * (.0092@107.53) = 1.4628@17.53 or 1.394 + .440j
Taking the real part of 1.394 + .440j means that when the voltage
source is at .707 volts, the voltage across L is at 1.394 volts.
------
Any comments or corrections would be greatly appreciated. Thanks