# Transients in RC and RL circuits analyzed using a Ti-84 Plus

On this page, the TI-84 Basic program for solving first-order differential equations is applied to two well-known electric circuits: the RC and RL circuit.

Besides, the YouTube movies, an extended example of the input behavior of a scope in DC or AC modes is demonstrated, showing the power of the TI-Basic programs. If you have no experience with numerical programs, it is advised to study the next page first;

## The RC circuit is supplied by a DC Voltage source

The RC series circuit is a well-known circuit in the world of electrical engineering. The RC-time constant, along with the charge curve of the capacitor, is a much used term in electrical engineering. In a YouTube movie, the program is explained

## Calculating inrush currents of a RL circuit supplied with a DC voltage

The RL circuit is an important circuit in electric engineering. The first order differential equation solver is applied and explained in the next YouTube video.

## Using AC or DC modes on a scope ?

With the help of the TI-Basic program, first order diff. eq. solver, we show you the differences between AC and DC modes on a scope. This example is chosen because it shows the perfect interaction between the available programs.

The input impedance of a scope in DC modes is typically 1 M Ω in practice. To separate an AC signal from a DC signal, the AC modes can be used, which means that a couple capacitor is put into series with the input. The schemes are shown here below. However, for low-frequency signals, this capacitor significantly influences the input signal, as we will see.

In DC modes, it yields that the scope voltage U0=Ui, so there is disturbance on Ui

However, in AC modes it yields:

Uo=Ui -Uc1

and

(R1C1)dUc1+Uc=Ui

dt

For a square wave of 20 V for Ui with a frequency of 1 kHz and 10 Hz, the voltage Uo is calculated using the Ti84. The capacitor value is 0.1uF and the input resistance is M Ω. The calculation is done in 3 steps:

1 : make a square wave as input with the program FUNCST1

2 : Solve the differential equation with the program DV1TRP2.8xp, which solves for Uc.

3: Calculate Uo using the program POSTPRE2.8xp, which gives Uo-Ui-Uc1

To make a square wave with more than 3 waves, we use the program FUNCST1.8xp instead of FUNCGEN.8xp; the function is stored in Y6.

## Step 1: the square wave

## Step 2 : solve the differential equation with the program DV1ETRP2.8xp

The program DV1ETRP2.8XP solves :

a dy +by=f(x) with y=Uc1, a=R1C1=0.1, b=1, f(x) =Ui

dx

The differential equation solver solves Uc. For 1 kHz square wave, the voltage Uc is very small (maximum 0.1V) This means that the voltage drop over the capacitor is very small and Uo is almost the same as Ui. Calculating Uo is the next step 3.

## Step 3 calculate Uo=Ui-Uc with the program POSTPRE2.8xp

The results of the differential equation solver can be post processed with the program POSTPRE2'. From the numerical results (which is y=Uc) it calculates

ady +by+cf(x). Which we can read as adUc1+bUc1+c(Ui)

dx dt

For a=0 b=-1 and c =1 : Ui-Uc1=Uo is calculated. See the results.

#### Conclusion: High-frequency signals can be reliably measured in AC modes.

## Same analysis with square wave input function of 10 Hz

Step 1 : create a square wave of 10 Hz (period time 0.1 sec) with FUNCST1.8xp

Step 2 : solve the differential equation with DV1ETRP2, time step 0.1/50 sec, a=0.1, calculate Uc

Step 3 : calculate Uo with POSTPRE2

See the results

#### Conclusion : Low frequency signals cannot be measured reliably in AC modes

The Results from the Ti-84 are in agreement with the results of Multisim

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