Finding Resistance of Voltmeter in RC Circuit

In summary: Sorry, is there a way you can post what the circuit looks like? A sketch or a scan? And is the AC nature of the circuit (you need AC if you have a capacitor) from an AC source, or a DC source with a switch closed at time...
  • #1
platonic
39
0
The question asks to derive the equation you need to find the resistance of a voltmeter in a simple RC circuit like the one shown here:

http://people.ee.duke.edu/~cec/final/node32.html

But R2 in the link is actually a capacitor in my circuit.
The equations in the lab journal are

Vc=q/C

V0=VR+Vc

I=VR/R

VC=V0(1-e^(-t/RC))

I=I0e^(-t/RC)

Vc=V0e^(-t/RC)

t=RCln(V0/VC)

VC=(Vm(R1+Rm))/RmI don't even know where to begin with this problem, because I'm not sure which equation to derive, let alone how to derive it. I'm guessing it's the last one I posted, but I'm not sure if the equation we are supposed to use was listed or not. Please help.
 
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  • #2
If the last equation is the one I need to derive, I'm guessing it has something to do with Ohm's law since there are Vs and Rs in it. I know the rules here are that I must attempt the problem, but I don't know wjere to start here.
 
  • #3
anyone?
 
  • #4
platonic said:
The question asks to derive the equation you need to find the resistance of a voltmeter in a simple RC circuit like the one shown here:

http://people.ee.duke.edu/~cec/final/node32.html

But R2 in the link is actually a capacitor in my circuit.



The equations in the lab journal are

Vc=q/C

V0=VR+Vc

I=VR/R

VC=V0(1-e^(-t/RC))

I=I0e^(-t/RC)

Vc=V0e^(-t/RC)

t=RCln(V0/VC)

VC=(Vm(R1+Rm))/Rm


I don't even know where to begin with this problem, because I'm not sure which equation to derive, let alone how to derive it. I'm guessing it's the last one I posted, but I'm not sure if the equation we are supposed to use was listed or not. Please help.

Welcome to the PF. I'm not sure I understand -- why was R2 changed to a capacitor? To make you derive the resistance of the voltmeter in a different way? So the source at the left Vs is now an AC generator, not the DC one shown?
 
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  • #5
berkeman said:
Welcome to the PF. I'm not sure I understand -- why was R2 changed to a capacitor? To make you derive the resistance of the voltmeter in a different way? So the source at the left Vs is now an AC generator, not the DC one shown?

Thanks for the welcome. I just posted the link to give you an idea of how the circuit I'm talkin about looks like. The R2 in the link is actually a capacitor in the circuit I'm asking about. And now that I'm looking at it, R1 is actually on the right side/loop/rectangle, not the left.

So R2 in the figure is a capacitor in my circuit, and R1 is on the right side of the dividing line, not the left. The small rectangle to the right has R1 on the top side, a capacitor on the left side, and the voltmeter on the right. And we are asked to find and derive the equation for the resistance of the voltmeter.
 
  • #6
platonic said:
Thanks for the welcome. I just posted the link to give you an idea of how the circuit I'm talkin about looks like. The R2 in the link is actually a capacitor in the circuit I'm asking about. And now that I'm looking at it, R1 is actually on the right side/loop/rectangle, not the left.

So R2 in the figure is a capacitor in my circuit, and R1 is on the right side of the dividing line, not the left. The small rectangle to the right has R1 on the top side, a capacitor on the left side, and the voltmeter on the right. And we are asked to find and derive the equation for the resistance of the voltmeter.

I'm not so worried about left-right. So you have an AC signal generator driving through R1 into the parallel combination of the voltmeter and a capacitor?
 
  • #7
I think R1 and the capacitor are in parallel. It looks like this

http://www.teslaradio.com/pages/compare_files/image021.gif

But the capacitor is on the line in the middle, the one with the orange dot, and the voltmeter is where the capacitor is in the link. That switch wasn''t there though.
 
  • #8
platonic said:
I think R1 and the capacitor are in parallel. It looks like this

http://www.teslaradio.com/pages/compare_files/image021.gif

But the capacitor is on the line in the middle, the one with the orange dot, and the voltmeter is where the capacitor is in the link. That switch wasn''t there though.

Sorry, is there a way you can post what the circuit looks like? A sketch or a scan? And is the AC nature of the circuit (you need AC if you have a capacitor) from an AC source, or a DC source with a switch closed at time t=0?
 
  • #9
This is the sketch. Should have done this at first. The source is DC with switch closed at t=0.
 

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  • #10
platonic said:
This is the sketch. Should have done this at first. The source is DC with switch closed at t=0.

Thanks, that helps. So just write the two KCL equations for the nodes and solve for v(t) as a function of the resistance of the DVM.
 
  • #11
By the two KCL equations I assume you mean the current laws for each node? And v(t) as a function of the resistance, I don't understand what you mean. I guess v(t) is voltage as a function of time, but how can you take a voltage as a function of time as a function of resistance? Like (v(t))(R)? BTW thanks for the help.
 
  • #12
platonic said:
This is the sketch. Should have done this at first. The source is DC with switch closed at t=0.

Actually, after further review, that circuit looks incorrect. You don't close a switch between an ideal voltage source and a capacitor -- that gives you infinite current. The first resistor would have to be between the switch and the capacitor, and the voltmeter would then be in parallel with the capacitor. The finite impedance of the voltmeter then alters the charging time constant of the circuit.
 
  • #13
berkeman said:
Actually, after further review, that circuit looks incorrect. You don't close a switch between an ideal voltage source and a capacitor -- that gives you infinite current. The first resistor would have to be between the switch and the capacitor, and the voltmeter would then be in parallel with the capacitor. The finite impedance of the voltmeter then alters the charging time constant of the circuit.

The source was open for only on for a small amount of time I guess to charge the capacitor, then it was cut off. The capacitor charged almost instantly, then it released it's charge slowly, and we measured the voltage across the voltmeter in 4 second intervals.

BTW now I am completely sure that the equation we must derive is the last one I listed. Vc=Vm((R1+Rm)/Rm).

(R1+Rm)/Rm is called alpha and it is the constant that describes the relationship between Vc and Vm. Because of R1, Vm will always differ from Vc by the constant alpha. This seems like Kirchoffs current law will come into play, correct? And I must do something witht he equations at the two nodes as you said before?
 
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  • #14
It says the equation can be derived using Ohm's law, but Vc is not stated in terms of V and R but, it's Vc=q/C and Vc=V0(1-e^(-t/RC)). There's no q, C or e in the equation.
 
  • #15
platonic said:
It says the equation can be derived using Ohm's law, but Vc is not stated in terms of V and R but, it's Vc=q/C and Vc=V0(1-e^(-t/RC)). There's no q, C or e in the equation.

I'm still not sure I understand the setup, but that equation is correct for a capacitor in parallel with the series combination of R1 and Rm. The "R" is the series combination of the resistances, and Vo is the initial voltage at t=0 when the switch is opened.
 
  • #16
berkeman said:
I'm still not sure I understand the setup, but that equation is correct for a capacitor in parallel with the series combination of R1 and Rm. The "R" is the series combination of the resistances, and Vo is the initial voltage at t=0 when the switch is opened.

So the setup of the derivation is Vc=V0/(R1+Rm)? It says to use Ohm's law for the derivation but that the Vc can't be expressed with Ohm's law? Or is it like IVc=I(R1+Rm)?
 
  • #17
platonic said:
So the setup of the derivation is Vc=V0/(R1+Rm)? It says to use Ohm's law for the derivation but that the Vc can't be expressed with Ohm's law? Or is it like IVc=I(R1+Rm)?

Well, what equation do you use to relate I(t) and V(t) for a capacitor? It's not Ohm's law -- that's for the resistive part of the circuit.
 
  • #18
Vc=V0(1-e^(-t/RC)) I presume. So I guess you replace V0 with IR to make it give I(t) as well, is this right? But this is a natural log so I'm not so sure. None of the equations they give have both I and V in them.
 
  • #19
platonic said:
Vc=V0(1-e^(-t/RC)) I presume. So I guess you replace V0 with IR to make it give I(t) as well, is this right? But this is a natural log so I'm not so sure. None of the equations they give have both I and V in them.

The current/voltage relation for a capacitor is:

[tex]I(t) = C \frac{dV(t)}{dt}[/tex]

You use this in the KCL equation(s) for the circuit to express how current and voltage are related for the whole circuit. You then assume a solution (involving the exponential decay) for the equations, and solve them based on the initial conditions that you are given (switch opened at t=0).
 
  • #20
berkeman said:
The current/voltage relation for a capacitor is:

[tex]I(t) = C \frac{dV(t)}{dt}[/tex]

You use this in the KCL equation(s) for the circuit to express how current and voltage are related for the whole circuit. You then assume a solution (involving the exponential decay) for the equations, and solve them based on the initial conditions that you are given (switch opened at t=0).

Okay thanks I think I should be able to do it now.
 

1. How do you find the resistance of a voltmeter in an RC circuit?

To find the resistance of a voltmeter in an RC circuit, you will need to use Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage can be measured by the voltmeter, and the current can be measured by an ammeter. Simply divide the voltage reading by the current reading to calculate the resistance of the voltmeter in the RC circuit.

2. Why is it important to find the resistance of a voltmeter in an RC circuit?

Knowing the resistance of a voltmeter in an RC circuit is important because it allows you to accurately measure the voltage across the circuit. Without taking into account the resistance of the voltmeter, the voltage reading could be inaccurate and lead to incorrect calculations and conclusions about the circuit.

3. Can the resistance of a voltmeter in an RC circuit change?

Yes, the resistance of a voltmeter in an RC circuit can change depending on the type of voltmeter being used. Analog voltmeters typically have a higher resistance compared to digital voltmeters, which have lower resistance. Additionally, the resistance of a voltmeter can also vary depending on the range setting being used.

4. How does the resistance of a voltmeter affect the accuracy of the voltage reading?

The resistance of a voltmeter can affect the accuracy of the voltage reading by creating a voltage divider effect in the circuit. This means that some of the voltage drops across the voltmeter instead of the rest of the circuit, resulting in a lower voltage reading. To minimize this effect, it is important to use a voltmeter with a high input impedance, as it will draw less current from the circuit and have a smaller impact on the voltage reading.

5. Are there any limitations to finding the resistance of a voltmeter in an RC circuit?

One limitation to finding the resistance of a voltmeter in an RC circuit is that it assumes the circuit is in a steady state. If the circuit is still charging or discharging, the resistance of the voltmeter may not accurately reflect the true resistance in the circuit. Additionally, this method may not work for more complex circuits with multiple resistors and capacitors.

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