Capacitor question.

[Soontir C'boath]

Tis a bit embarrassing and I'm tired right now to boot but I have a lab report that needs to be done by tomorrow. The lab used an oscilloscope, a battery, a switch (to disconnect from the battery), and two capacitors one of which has an unknown capacitance. Basically, I plotted a graph where the y axis is the observed voltage over initial voltage and the x axis with the decade box's varying capacitance.

The lab says to use the graph to find the value of the unknown capacitance. The question is how do I do it correctly? I think I will most likely feel stupid after I read the method but me sludgy brain needs some Zs. Thanks in advance.

[SWPIGWANG]

The circuit probably have a resistor (or at least resistance). Find the RC time constant from the graph and figure out the cap?

[muse]

Would you happen to have a schematic of the lab setup?

[Superman]

Are we talking about a Flux Capacitor by any chance? If so, boy do I have some plans for that... Look out Karate Kid part 1! I'll be watching you in the original theatrical release!

[Howedar]

Tis a bit embarrassing and I'm tired right now to boot but I have a lab report that needs to be done by tomorrow. The lab used an oscilloscope, a battery, a switch (to disconnect from the battery), and two capacitors one of which has an unknown capacitance. Basically, I plotted a graph where the y axis is the observed voltage over initial voltage and the x axis with the decade box's varying capacitance.

The lab says to use the graph to find the value of the unknown capacitance. The question is how do I do it correctly? I think I will most likely feel stupid after I read the method but me sludgy brain needs some Zs. Thanks in advance.

I dunno, I'd think you'd look at the maximum power delivery of the battery and compared that to d/dt (1/2 C V^2)

[Fingolfin_Noldor]

Capacitance is given by C = Q / V where Q is the charge and V the potential difference across the two plates. The question of course is what your graph is as the "observed voltage" is kind of vague. What is your circuit setup?

[Fingolfin_Noldor]

Ghetto Edit: A possible way of calculating capacitance is plotting Q vs V and getting the gradient of the graph. If you have dC/dt = I/V, then perhaps finding the gradient at a point and then use the equation and integrating both sides to get the capacitance at that point.

[Wyrm]

Okay, let's assume that when you flip the switch, the capacitor is connected directly to some load with some unknown but constant resistance R. As stated, you have a capacitor of unknown capacitance (C_t) and and capacitor of known capacitance (C_r). Initially, the voltage across the terminals of the capacitor will be V_0, and it drops to V_1 as time goes from 0 to T.

From the definition of capacitance, Q = CV, and Ohm's law has V = IR. Now, I = -dQ/dt in this case, so take the time derivative of the first equation to get dQ/dt = C dV/dt (C is constant), and make a substitution in Ohm's law to get V = -CR dV/dt.

Separate the variables: dV/V = -dt/CR

Integrate both sides: ln V = -t/CR + c

Rearrange the equation: V = V_0 exp(-t/CR)

Therefore, the graph should look like an exponential decay curve.

Remember that we know one of the capacitances, C_r. Going back to dV/V = -dt/CR, and integrating the left side from V_0 to V_1 and the right side from 0 to T, we get ln(V_1/V_0) = -T/CR, or alternately, CR = ln(V_0/V_1)/T. You can then use this equation to derive the unknown resistance from the known capacitance, and therefore derive the unknown capacitance from the now-known resistance.

[Wyrm]

GHETTO EDIT: Er, the last equation should be CR = T/ln(V_0/V_1). Sorry for the screwup.

[Soontir C'boath]

Wyrm, I'm not sure that's what they're asking us to do but I'll keep that in mind especially when Ohm's Law is next week.

Capacitance is given by C = Q / V where Q is the charge and V the potential difference across the two plates. The question of course is what your graph is as the "observed voltage" is kind of vague. What is your circuit setup?

It's the first diagram that can be found here

I should've said it was a double throw switch in which one position would connect the battery to the known capacitor. When the switch is thrown to the other position, the battery is disconnected and the known capacitor is parallel to the unknown and discharges to it. The oscilloscope reads the voltage drop in the known capacitor. We're suppose to do this with five different capacitance levels that we can set on the capacitor (it's a decade cap. box).

In the graph, it'd be V_recorded/V_original on the y-axis and the x-axis is the microfarads that the known capacitor was set. V_recorded being the voltage drops observed when the known capacitor discharges.

[Wyrm]

Wyrm, I'm not sure that's what they're asking us to do but I'll keep that in mind especially when Ohm's Law is next week.

Oh. Actually, I would have thought you'd learn Ohm's law first.

It's the first diagram that can be found here

I should've said it was a double throw switch in which one position would connect the battery to the known capacitor. When the switch is thrown to the other position, the battery is disconnected and the known capacitor is parallel to the unknown and discharges to it. The oscilloscope reads the voltage drop in the known capacitor. We're suppose to do this with five different capacitance levels that we can set on the capacitor (it's a decade cap. box).

In the graph, it'd be V_recorded/V_original on the y-axis and the x-axis is the microfarads that the known capacitor was set. V_recorded being the voltage drops observed when the known capacitor discharges.

Ah. See, this is why circuit diagrams are important. The setup described here does not match my vision of the circuit diagram at all. I was envisioning a setup where C1 was actually a socket where you could swap out the reference (known) capacitor with the test (unknown) capacitor and C2 in the diagram was actually a fixed resistor (load).

After charging the reference capacitor, C1 (with unknown capacitance C_1), that capacitor has a charge on it Q_{total} = C_1 V_{initial}, and the test capacitor, C2 (with known capacitance C_2), is completely discharged.

After you throw the switch, the charge rearranges itself so that the voltage across both capacitors is identical, V_{final}, and the oscilliscope measures the drop V_{initial}/V_{final}. V_{final} in both capacitors C1 and C2 implies that Q_1 = C_1 V_{final} and Q_2 = C_2 V_{final}, where Q_1 and Q_2 are the charges on the unknown and known capactiors' plates, respectively. Rearranging both equations, we get V_{final} = Q_1/C_1 = Q_2/C_2. But Q_1 and Q_2 came from the original charge on the test capacitor, Q_{total}; Q_1 + Q_2 = Q_{total}, so Q_1 + Q_1 C_2/C_1 = Q_{total} = Q_1 (1 + C_2/C_1), or Q_{total}/Q_1 = 1 + C_2/C_1. But Q_{total} = C_1 V_{initial} and Q_1 = C_1 V_{final}, therefore, V_{initial}/V_{final} = 1 + C_2/C_1.

From here, it's easy to answer Q1. Q2 comes from the realization that the circuit is not perfect.

For Q3, if we vary C_2 and look at V_{initial}/V_{final}, it turns out that this is a line: d(V_i/V_f) = 1/C_1 d(C_2), or d(V_i/V_f)/d(C_2) = 1/C_1 = a constant depending only on C_1. The slope of this line is the reciprocal of the unknown capacitor's capacitance.