# I'm clueless. Oscilloscope as resistor

1. Mar 21, 2004

### KingJaymz

I've thought and thought and thought about this, but I can't figure it out. How might I be able to estimate the apparent resistance of an oscilloscope? We're dealing with an oscilloscope connected to a circuit which uses a resistor and a capacitor. I think this question might have something to do with it. What function does the resistor play in a circuit once the voltage source is removed?

2. Mar 22, 2004

### paul11273

Can you be a little more specific as to how the power source, resistor and capacitor are wired together?
Is the resistor in series or parallel with the capacitor?

3. Mar 22, 2004

### KingJaymz

Yeah. Shoulda included that earlier. I was hoping it was more of a conceptual question. We have a 1.5 V battery powering a circuit on a breadboard that's got a 47 kiloohm resistor and a 33 microfarad capacitor in it. The resistor and capacitor are in series and the test leads of the oscilloscope are connected to the end of the breadboard circuit and the negative end of the battery. The oscilloscope completes the circuit. Did I explain that well?

4. Mar 22, 2004

### paul11273

OK, so when you remove the power source, the cap is discharging through the O-scope, right?

If that is the case, you can monitor the voltage on the scope as it varies with time. The voltage will be decaying.

You can find the resistance by solving this equation for R:

v(t)=V*e^(-t/RC)
where:
v(t) is the voltage at a given time
V is the initial voltage, 1.5 volts in your case
t is the particular time you are reading the voltage v(t)
C is the capacitance of the capacitor
R is the resistance of the O-scope, for which you are solving.

Solve this equation for R (hint: you will have to take the ln of both sides to eliminate e) and you will have general equation with which you will perform your calculations.

I would take "snapshots" of the voltage at specific time increments, say maybe every 5 seconds, and come up with a table.
For example:

t=0 , v(t)=1.5v
t=5 , v(t)=.900v
t=10 , v(t)=.700v

Monitor the voltage for about 1 minnute, that should give enough data.

Plug these values into the equation you found, and calculate R at each one of these data sets of t and v.

Average the values, and you will have a good approximation of resistance of the scope.

Confusing? Let me know, I will try to clarify.

5. Mar 22, 2004

### Chen

Just a correction, R is the the resistance of the oscilloscope + the resistance of your resistor ($47K\Omega$).

6. Mar 22, 2004

### paul11273

Only if the cap is discharging through both. If he puts the o-scope directly across the cap, it will give only the R of the scope.

I wasn't sure by the description of the setup which was the case.

Other than that, you agree with this method? I wonder if there is a shorter way to figure this out. Any ideas?

7. Mar 22, 2004

### Chen

He only said that "the voltage source is removed" so I thought the original resistor is not removed.

But yes, I agree with your method and that's how I'd do it. The easiest way is to connect the scope directly to the source, or to an ohmmeter, but I don't think that's allowed.

8. Mar 22, 2004

### KingJaymz

Ah, I see. Thank you all very much. That was very helpful.