J
[email protected]
- Jan 1, 1970
- 0
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 [email protected]
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 / [email protected] = [email protected] 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) * ([email protected]) = [email protected] 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) * ([email protected]) = [email protected] 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) * ([email protected]) = [email protected] 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.
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 [email protected]
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 / [email protected] = [email protected] 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) * ([email protected]) = [email protected] 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) * ([email protected]) = [email protected] 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) * ([email protected]) = [email protected] 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.