TI-55 III Programs Part II: Impedance of a Series RLC
Circuit, Quadratic Equation, Error Function
On to Part II!
For Part I, click here: Digital Root, Complex Number Multiplication, Escape Velocity
For Part III, click here: Area and Eccentricity of Ellipses, Determinant and Inverse of 2x2 Matrices, Speed of Sound/Principal Frequency
TI-55 III: Impedance
of Series RLC Circuit
The impedance of a series RLC circuit in Ω (ohms) is:
Z = √(R^2 + (2*π*f*L – 1/(2*π*f*C))^2)
Where:
R = resistance of the resistor in ohms (Ω)
L = inductance of the inductor in Henrys (H)
C = capacitance of the capacitor in Farads (F)
f = resonance frequency in Hertz (Hz)
XL = 2*π*f*L
XC = 1/(2*π*f*C)
Program:
Partitions allowed: 1-4
STEP
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CODE
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KEY
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COMMENT
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00
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65
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*
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Start with f
|
01
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02
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2
|
|
02
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65
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*
|
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03
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91
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π
|
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04
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95
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=
|
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05
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61
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STO
|
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06
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00
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0
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Store 2πf in R0
|
07
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65
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*
|
|
08
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12
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R/S
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Prompt for L
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09
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75
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-
|
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10
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53
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(
|
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11
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71
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RCL
|
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12
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00
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0
|
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13
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65
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*
|
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14
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12
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R/S
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Prompt for C
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15
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54
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)
|
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16
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17
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1/x
|
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18
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18
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X^2
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19
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85
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+
|
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20
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12
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R/S
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Prompt for R
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21
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18
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X^2
|
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22
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95
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=
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23
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13
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√
|
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24
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41
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INV
|
|
25
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47
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Eng
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Cancel Eng Notation
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26
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12
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R/S
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Display Z
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Input: f [RST] [R/S],
L [R/S], C [R/S], R [R/S]
Result: Z
Test:
f = 60 Hz
L = 0.25 H
C = 16 * 10^-6 F
R = 150 Ω
Result: 166.18600 Ω
TI-55 III: Quadratic Equation
This program find the real roots of the equation:
X^2 + B*X + C = 0
Where:
D = B^2 – 4*C
If D ≥ 0, then continue the program since it will find
the real roots. Otherwise, stop since
the roots are complex and is beyond the scope of this program. The two roots are:
X1 = (-B + √D)/2
X2 = (-B - √D)/2
Program:
Partitions Allowed: 3
STEP
|
CODE
|
KEY
|
COMMENT
|
00
|
71
|
RCL
|
Calculate
Discriminant
|
01
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00
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0
|
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02
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18
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X^2
|
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03
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75
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-
|
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04
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04
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4
|
|
05
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65
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*
|
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06
|
71
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RCL
|
|
07
|
01
|
1
|
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08
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95
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=
|
|
09
|
12
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R/S
|
Display
Discriminant
|
10
|
13
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√
|
|
11
|
61
|
STO
|
|
12
|
02
|
2
|
|
13
|
75
|
-
|
|
14
|
71
|
RCL
|
|
15
|
00
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0
|
|
16
|
95
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=
|
|
17
|
55
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÷
|
|
18
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02
|
2
|
|
19
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95
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=
|
|
20
|
12
|
R/S
|
Display X1
|
21
|
53
|
(
|
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22
|
71
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RCL
|
|
23
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00
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0
|
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24
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85
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+
|
|
25
|
71
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RCL
|
|
26
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02
|
2
|
|
27
|
54
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)
|
|
28
|
94
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+/-
|
|
29
|
55
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÷
|
|
30
|
02
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2
|
|
31
|
95
|
=
|
|
32
|
12
|
R/S
|
Display X2
|
Input: B [STO] 0,
C [STO] 1, [RST] [R/S]
Results:
Discriminant [R/S], root 1 [R/S], root 2
Test: Solve X^2 +
0.05*X – 1 = 0
Input: 0.05 [STO]
0, 1 [+/-] [STO] 1 [RST] [R/S]
Results:
Discriminant = 4.0025 (It is non-negative,
continue) [R/S]
X1 ≈ 0.9753125
[R/S]
X2 ≈ -1.0253125
TI-55 III: Gaussian Error Function
The error function is defined as:
erf(x) = ∫( 2*e^(-t^2)/√π dt, from t = 0 to t = x)
This program illustrates the integration function [ ∫ dx
] on the TI-55 III.
Program:
Prepare by pressing [2nd] [LRN] (Part) 3. Integration needs a minimum of 3 memory
registers. That means, f(x) can take a
maximum of 40 steps.
STEP
|
CODE
|
KEY
|
COMMENT
|
00
|
18
|
X^2
|
Integrand
|
01
|
94
|
+/-
|
|
02
|
41
|
INV
|
|
03
|
44
|
ln x
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[INV] [ln x]:
e^x (EXP)
|
04
|
65
|
*
|
|
05
|
02
|
2
|
|
06
|
55
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÷
|
|
07
|
91
|
π
|
|
08
|
13
|
√
|
|
09
|
95
|
=
|
|
10
|
12
|
R/S
|
|
11
|
22
|
RST
|
End f(x) with
=,R/S,RST
|
Input: 0 [STO] 1
(lower limit), x [STO] 2 (upper limit), [ ∫ dx ] n (number of partitions) [R/S]
Result: erf(x)
Test 1: erf(1.2) ≈ 0.910314. I use 12 partitions.
Input: 0 [STO] 1,
1.2 [STO] 2, [ ∫ dx ] 12 [ R/S ]
Result: 0.910314
Test 2: erf(0.9) ≈
0.7969082. Store 0 in R0, 0.9 in
R1. 12 partitions are used.
This blog is property of Edward Shore, 2016.