Capacitive Network: Calculating Voltage between Two Capacitors

[Anil Tanwar]

Hi all,
I am new to this group and joined after reading this interesting discussion. In extention th the David, I've one doubt to find the voltage between two capa. Suppose a capa network is like Vsupply(1V)-C1-C2-gnd; Let say Vx is the between node name. Vx1=C1/(C1+C2)*1v =0.5v if C1=C2; Now Vsupply=2v, I want to calculate new Vx1, say Vx2, from charge-conservation theory.Previous total charge was 1v*[C1*C2/(C1+C2)] -series combination; Current total charge is (2v-Vx2)C1 +Vx2*C2; If you equate both charge, Vx2=1/(C2-C1)*[1v*{C1*C2/(C1+C2)}-2v*C1] which infinity if C1=C2.
Pls you (Russ or anybody) point out where I am wrong; Russ

 

[Astronuc]

Previous total charge was 1v*[C1*C2/(C1+C2)] -series combination; Current total charge is (2v-Vx2)C1 +Vx2*C2; If you equate both charge,

One cannot equate both charges. If one doubles the voltage, the charge on the capacitors changes, in fact it doubles, and the charge on each capacitor doubles and is proportionally distributed as before.

Where there are two elements in series, there is a voltage drop across each.

Given a V --| C1 |-----| C2 |----- gnd, V = V1 + V2, but capacitors in series are treated like resistors in parallel by virtue of the relationship between charge and voltage. Current is a transient property in capacitors, since when a voltage is applied, current flows until a new equilibrium voltage (charge) is achieved.

The combined capacitance of C1 and C2 is given by,

1/C = 1/C1 + 1/C2, or C = (C1*C2)/(C1+C2). If C1 = C2, then C = (C1*C1)/(C1+C1) = (C1)/2

And V = Q/C => V = Q (C1+C2)/(C1*C2), and by charge conservation, the charge is the same on each capacitor, and the net charge in a capacitor is zero, i.e. the negative charge on one plate is balanced by the same magnitude of positive charge on the other plate.

So Q = Q1 = Q2 or C*V = C1*V1 = C2*V2, so one can find V1 or V2 in terms of V,

V1 = (C*V)/C1 = [(C1*C2)/(C1+C2) / C1] V = C2/(C1+C2) V, and likewise
V2 = C1/(C1+C2) V.

When the voltage is change, the charge on each capacitor changes simultaneously.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html

 

[kyle8921]

Hi Russ,

Thank you for sharing your question and calculations with us. It looks like you are on the right track with using charge conservation theory to calculate the voltage between two capacitors in a capacitive network. However, there are a few points that I would like to clarify.

Firstly, when you say Vx1=C1/(C1+C2)*1v =0.5v if C1=C2, I believe you meant to say Vx1=C1/(C1+C2)*2v =1v if C1=C2, since the supply voltage is now 2v.

Secondly, in your equation for Vx2, you have a typo where you wrote "1v*{C1*C2/(C1+C2)}" instead of "2v*{C1*C2/(C1+C2)}". This should be corrected to accurately represent the total charge.

Lastly, your calculation for Vx2 will not result in infinity if C1=C2. In fact, it will result in Vx2=1v, which makes sense since in this case the capacitors are essentially acting as one capacitor with double the capacitance. Therefore, the voltage across each capacitor will be equal to half of the supply voltage, which is 1v.

I hope this helps clarify your calculations. Keep up the good work and happy experimenting!

Best,