TI-30Xa Algorithms: RLC Series Circuit
The
task is to calculate the total impedance and phase angle for an RLC
circuit in a series. An RLC circuit contains a resistor ( R ), an
inductor ( L ), and a capacitor ( C ) , all powered by a voltage
supply. The resistor, inductor, and capacitor are all in a single
path.
The
three elements are measured in the following units:
Resistor
( R ): Ohms ( Ω )
Inductor
( L ): Henry ( H )
Capacitor
( C ): Farads ( F )
The
impedance for the RLC circuit is calculated by:
Z =
√( R^2 + ( (2 * π * f * L) - 1 / (2 * π * f * C) )^2 )
The
phase angle is calculated by:
θ =
arctan( ( (2 * π * f * L) - 1 / (2 * π * f * C) ) / R )
where:
f = frequency in Hertz (Hz)
We
can use the rectangular to polar conversion function (R>P) to
calculate both the impedance and phase angle.
x =
R
y =
(2 * π * f * L) - 1 / (2 * π * f * C)
If
we have an RL circuit, with no capacitor, then C = 0.
If
we have an RC circuit, with no inductor, then L = 0.
If
we have an LC circuit, with no resistor, then R = 0.
TI-30Xa
Algorithm – RLC Circuit
Step
1: Store resistance in Memory 1.
R [
STO ] 1
Step
2: Calculate angular frequency, ω = 2 * π * f, store in memory 3.
2 [
× ] [ π ] [ × ] f [ = ] [ STO ] 3
Step
3: Calculate (2 * π * f * L) - 1 / (2 * π * f * C)
[
RCL ] 3 [ × ] L [ - ] [ ( ] [ RCL ] 3 [ × ] C [ ) ] [ 1/x ] [ = ] [
STO ] 2
If
we have an RL circuit, Steps 2 and 3 can be shortened to:
2 [
× ] [ π ] [ × ] f [ × ] L [ = ] [ STO ] 2
If
we have an RC circuit, Steps 2 and 3 can be shortened to:
[ (
] 2 [ × ] [ π ] [ × ] f [ × ] C [ ) ] [ 1/x ] [ +/- ] [ = ] [
STO ] 2
Step
4: Calculate Impedance and Phase Angle:
[
RCL ] 1 [ 2nd ] [ π ] (x<>y) [ RCL ] 2 [ 2nd
] [ - ] (R>P)
What
is shown: (r) Impedance
Press
[ 2nd ] [ π ] (x<>y) for phase angle (θ)
Examples
Example
1: RLC Series Circuit
R =
50 Ω
L =
3.8 H
C =
0.7 F
f =
40 Hz
Step
1: Store resistance in Memory 1.
50 [
STO ] 1
Step
2: Calculate angular frequency, ω = 2 * π * f, store in memory 3.
2 [
× ] [ π ] [ × ] 40 [ = ] [ STO ] 3 (251.3274123)
Step
3: Calculate (2 * π * f * L) - 1 / (2 * π * f * C)
[
RCL ] 3 [ × ] 3.8 [ - ] [ ( ] [ RCL ] 3 [ × ] 0.0007 [ ) ] [ 1/x ]
[ = ] [ STO ] 2
(949.36000616)
Step
4: Calculate Impedance and Phase Angle:
[
RCL ] 1 [ 2nd ] [ π ] (x<>y) [ RCL ] 2 [ 2nd
] [ - ] (R>P)
Impedance:
950.6758262
Phase
Angle: 86.9851854°
Example
2: RLC Series Circuit
R =
450 Ω
L =
1.15 H
C =
3.5 μF = 3.5 * 10^-6 F
f =
60 Hz
Step
1: Store resistance in Memory 1.
450
[ STO ] 1
Step
2: Calculate angular frequency, ω = 2 * π * f, store in memory 3.
2 [
× ] [ π ] [ × ] 60 [ = ] [ STO ] 3 (376.9911184)
Step
3: Calculate (2 * π * f * L) - 1 / (2 * π * f * C)
[
RCL ] 3 [ × ] 1.15 [ - ] [ ( ] [ RCL ] 3 [ × ] 3.5 [ EE ] 6 [ +/- ]
[ ) ] [ 1/x ] [ = ] [ STO ] 2
(-324.3408952)
Step
4: Calculate Impedance and Phase Angle:
[
RCL ] 1 [ 2nd ] [ π ] (x<>y) [ RCL ] 2 [ 2nd
] [ - ] (R>P)
Impedance:
554.7044405
Phase
Angle: -35.78246242°
Source
“Impedance
and Complex Impedance” Electronics Tutorials. AspenCore, Inc.
2024. Retrieved October 7, 2024.
https://www.electronics-tutorials.ws/accircuits/impedance.html
This
wraps up the TI-30Xa Algorithm Series. Next up will be a series on
the Casio fx-991CW.
I
want to wish you all a Happy New Year and a prosperous, sane, and
happy 2025! Be safe, everyone, it’s a very crazy world we live in.
Eddie
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