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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