# VCalculate the potential difference (in volts) between the terminals of the battery

I reduced to equivalence capacitance and found the potential difference between the nodes based on what was provided using the rules of parallel and series and my final answer doesn't match the answer given. I have a test in two days and have been studying this along with other subjects. Could someone please lay out the answer methodically for me?

This was from a practice test he gave us to do,which he gave last year. I've done the other question but am stuck on this one.

The answer btw
a) is 1.7V
b) is 1.3 C

## Homework Statement

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## Answers and Replies

TSny
Homework Helper
Gold Member

Hello, zeta2. According to the guidelines for this forum, you must show some work in order to receive help. You can click on the Rules tab at the top of the page to read the guidelines.

If you provide some detail in your attempt at solving the problem, we can help you find your mistake.

Delphi51
Homework Helper

It seems to me the potential across the 5.3 μC cap is V = Q/C = .5094.
Did you get that as a first step?
If the capacitors were initially uncharged, I think there is something about the 5.3 and 6.7 μC caps in series that must be the same, and allows you to calculate the voltage on it, too.

Let us see what you've got so far and then we can help more!

It seems to me the potential across the 5.3 μC cap is V = Q/C = .5094.
Did you get that as a first step?
If the capacitors were initially uncharged, I think there is something about the 5.3 and 6.7 μC caps in series that must be the same, and allows you to calculate the voltage on it, too.

Let us see what you've got so far and then we can help more!

What I did was first made an equivalence capacitor between the 5.3 and 6.7 C's they are in series so Ceq =(1/5.3 +16.7)^-1 =2.959. In that the charge for the 5.3 Cap was 2.7, then the 6.7 has to have the same, and the total charge for the Ceq capacitance would be 2.7. then I reduced the 3.3 and 2.959 capacitance, now in parallel to and new equivalent Ceq = 6.2516. The sum charge would be Qtotal = 2.7 +q(on 3.3 cap) = Ceq*V where Ceq = 6.2516 with some algebra I found the charge for q(on 3.3 cap) to be 4.133 and the V to be 1.0931....i think something's fuzzy with what I did though. Does everything I've done so far seem correct? I basically continued the same process untill it was down to an super (lol) eqv Capacitance (5.44 175) and the 1.8 in parallel.....Does anything I've done so far make sense? lol

TSny
Homework Helper
Gold Member

Looks like you're thinking about it ok, but your result 4.133 for the charge on the 3.3 cap and 1.0931 for the voltage V are not correct.

It might be a good idea to work out the voltages on each capacitor as you go along.

So, what do you get for V for the 5.3 and 6.7 capacitors individually?

Delphi51
Homework Helper

In that the charge for the 5.3 Cap was 2.7, then the 6.7 has to have the same,
Agree with that, and then you can get the V = Q/C = .4030 for the 6.7. The 5.3 and the 6.7 are in series so the voltage across the two of them is .4030 + .5094 = .9124
That same potential is across the 3.3 so you can work out its Q=CV = 3.011.
This puts us in disagreement! I don't understand how you got 4.133 for the 3.3's charge and 1.0931 for the Voltage.

I finished the problem and got 1.65 V without using any series or parallel capacitance formulas.

Oops, cross post TSny (I'm really slow) - but looks like we are giving the same advice!

Looks like you're thinking about it ok, but your result 4.133 for the charge on the 3.3 cap and 1.0931 for the voltage V are not correct.

It might be a good idea to work out the voltages on each capacitor as you go along.

So, what do you get for V for the 5.3 and 6.7 capacitors individually?

Yeah, that's what I did but I'm very sloppy with my notes. I'ma work this whole problem out again neatly when i'm done with these mechanics questions.

TSny
Homework Helper
Gold Member

Do you see how to get the voltage across the 3.3 capacitor from the voltages for the 5.3 and the 6.7 capacitors?

Do you see how to get the voltage across the 3.3 capacitor from the voltages for the 5.3 and the 6.7 capacitors?

Yeah, use the fact that the the sum of the charges =V*Ceq?

Edit:
I think I also see the error of my way! the next update shall be my complete answer!

Last edited:
TSny
Homework Helper
Gold Member

Yeah, use the fact that the the sum of the charges =V*Ceq?
It's easier to get the voltage across the 3.3 cap by just adding the voltages across the 5.3 and 6.7 capacitors. No need to find the equivalent capacitance to get the voltage across the 3.3 capacitor (as Delphi51 noted).

It's easier to get the voltage across the 3.3 cap by just adding the voltages across the 5.3 and 6.7 capacitors. No need to find the equivalent capacitance to get the voltage across the 3.3 capacitor (as Delphi51 noted).

Yeah, I just realize some of the things I've done are redundant and that's sorta piled to the point where I got caught under my own redundancy! LOl. However what I did was essentially what he did.

Agree with that, and then you can get the V = Q/C = .4030 for the 6.7. The 5.3 and the 6.7 are in series so the voltage across the two of them is .4030 + .5094 = .9124
That same potential is across the 3.3 so you can work out its Q=CV = 3.011.
This puts us in disagreement! I don't understand how you got 4.133 for the 3.3's charge and 1.0931 for the Voltage.

I finished the problem and got 1.65 V without using any series or parallel capacitance formulas.

Oops, cross post TSny (I'm really slow) - but looks like we are giving the same advice!

Yeah just did the same thing and got 1.65202 myself. :) just needed to keep my numbers in order!
Thank you guys! :)

for the charge on the 1.5 Cap, I got 1.40...my reasoning is sloppy...

Last edited:
Delphi51
Homework Helper

Congrats!
It was not productive to combine the capacitors because it obscured the ability to deduce the charges and voltages around the circuit.

Congrats!
It was not productive to combine the capacitors because it obscured the ability to deduce the charges and voltages around the circuit.

Yeah I just realize that was my problem and hence what just happened with the 1.5 Capcitence :(