RC circuit analysis using oscilloscopes

In summary, the conversation involved a practical examination where the goal was to find the values of an unknown resistor and capacitor using a signal generator and oscilloscope. The attempt at a solution involved comparing voltage drops and using Ohm's Law to obtain the magnitude of the impedance. The angle of the impedance was determined by comparing the phase difference between a purely resistive impedance and the RC impedance. The resulting values were represented in the form a + jb, where a is the value of the resistor and b is the value of the capacitor.
  • #1
blueincubi
1
0

Homework Statement



this was a practical examination given to us.


[connected to CH1]
|----(1.8k ohms)-----|
|aaaaaaaaaaaaaaaaaa|
|aaaaaaaaaaaaaaaaaa(unknown resistance) [these 2 unknowns are ch2]
(~)aaaaaaaaaaaaaaa|
|aaaaaaaaaaaaaaaaa(unknown capacitance)
|--------------------|

so you've got a signal generator which is covered up, and the components are hooked up to the oscilloscope in the manner described above. the problem was to find out the the values of the unknown resistor and capacitor using the graph in the oscilloscope.


2. Equations

t/T = x/360

where:
t is the horizontal distance between the two curves
T is period
x is the angle

Ohm's law

The Attempt at a Solution




what i tried to do was to compare the voltage drop between ch2 and ch1 (which i treated as two separate impedances). and then divided it by the known R to get a value I which is the current for the whole mesh(?).

using that value I, i obtained the magnitude of the impedance of the RC component by Ohm's Law.

for the angle of the impedance, i compared the phase difference between the pure R impedance and the RC impedance. this is the part I am not sure of, is the angle x i get be the angle of the RC impedance (since purely resistive impedances are 0degs)?

but then, i assumed that the above were correct and managed to get a form a + jb, where j is sqrt(-1), so i took it as the a being the value of the resistor and b as the value of the capacitor?
 
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  • #2
I don't understand your diagram.

Is it like this:

[PLAIN]http://dl.dropbox.com/u/4222062/R-R-C.PNG [Broken]

If so, maybe you could just refer to this diagram and say where the components connect to, or, you could modify the diagram using MS Paint and repost it?
 
Last edited by a moderator:

1. What is an RC circuit?

An RC circuit is a circuit that consists of a resistor (R) and a capacitor (C) connected in series or in parallel. It is used to filter and control the flow of electric currents.

2. How does an oscilloscope help with RC circuit analysis?

An oscilloscope is a device that measures and displays the voltage and current in a circuit in the form of a graph. It helps in analyzing the behavior of the RC circuit by showing the time-varying signals and the effects of the resistor and capacitor on the circuit.

3. What is the purpose of using an RC circuit in electronic devices?

An RC circuit can be used for various purposes, such as filtering out unwanted frequencies, smoothing out signals, and creating time delays. It is commonly used in electronic devices like filters, timers, and oscillators.

4. How do you calculate the time constant of an RC circuit?

The time constant (τ) of an RC circuit is calculated by multiplying the resistance (R) in ohms by the capacitance (C) in farads. It is expressed in seconds (s) and represents the time it takes for the capacitor to charge or discharge to 63.2% of its maximum value.

5. What are some common issues that can arise during RC circuit analysis?

Some common issues that can arise during RC circuit analysis include incorrect connections, damaged components, and unexpected signals on the oscilloscope. It is important to double-check connections and troubleshoot any issues before proceeding with the analysis.

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