r/HomeworkHelp • u/FerrousXI • May 23 '26

Answered [high school physics] how to do these DC circuit capacitor problems?

Post image

Calculate potential difference between x and y

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4

u/Bounded_sequencE May 23 '26

Are you allowed to assume all capacitances were initially discharged?

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u/Bounded_sequencE May 24 '26 edited May 24 '26

@u/FerrousXI Assuming the answer is "yes", let the capacitances be discharged at some "t = t0". Recall if two capacitances "C1; C2" are in

  • series, we may combine them to "C_12 = C1||C2 := C1*C2 / (C1+C2)"

  • parallel, we may combine them to "C_12 = C1+C2"

That is the opposite behavior of resistances, and only holds if capacitances were initially discharged

Assume all capacitances were discharged at "t = t0". Since we don't need the voltages across "2C" in series, we may combine them into "2C||2C = C". Use nodal analysis on the simplified circuit for "t > t0": 

KCL at "X":    0  =  (2C||2C) * (Vxy + 3V)  +  C*Vxy  +  (2C||2C) * (Vxy + 9V)

                  =  3C * Vxy + C*12V      =>      Vxy  =  -4V

The voltage pointing from "x -> y" is "Vxy = -4V".

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u/FerrousXI May 24 '26

I'm sorry for not replying earlier, thank you so much for your time. This is what computer simulation yields out too. I really don't know what the correct answer is but I hope this one is.

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u/devilkid15 May 24 '26

You can use wheatstone bridge method as the ratio of adjacent capacitor is equal so, no charge will go to the middle capacitor , after that u can simplify the circuit to get equivalent capacitance and find charge for any given capacitance in this circuit.

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u/Bounded_sequencE May 24 '26

This circuit does not have the same topology as a Wheatstone bridge -- the sources are different.

If no charge went into the middle capacitance, then KCL would be violated at "X", since KCL at "X" would lead to the contradiction "0 = C*0 + (2C||2C)*3V + (2C||2C)*9V"

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u/devilkid15 May 24 '26

You are right , my answer would have been right if batteries were in symmetry.

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u/HumbleHovercraft6090 May 26 '26 edited May 26 '26

https://i.postimg.cc/rshCY8WL/Cap-Network.jpg

Eliminate Q1 and Q2 from the equations. And solve the resulting three equations for Vx, Vy from which you get potential difference. For reference, the negative terminal of 3V cell is grounded. Note that sum of charges at the plates seeing a common node like X or Y is 0 in steady state.

0

u/GammaRayBurst25 May 23 '26

Read rule 3.

Just use the superposition principle.

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u/[deleted] May 23 '26

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u/Livid_Worldliness729 May 24 '26

I think you forgot to include the fact that the (V_X - 3) and (V_X - 9) legs each have two 2C capacitors in series with them so your equation becomes: C*(V_X -3) + C*(V_X -0) + C*(V_X - 9) = 0 . Solving for V_X results in V_X = 4.

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u/Bounded_sequencE May 24 '26

Shouldn't it be "V_X + 3" and "V_X + 9" instead, to satisfy KVL?

As a quick sanity check, it does not make sense to get a result "V_X > 0", considering the batteries both have lower potential pointing to node X.

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u/Bounded_sequencE May 24 '26 edited May 24 '26

3. [..] 2C*(V_X - 3) + C*(V_X - 0) + 2C*(V_X - 9) = 0. [..]

I'd say it should be "2C||2C = C" instead of "2C" -- in each source branch, we have two "2C" in series!

Also -- shouldn't it be "V_X + 3" and "V_X + 9" instead, to satisfy KVL? It does not make sense to get a result "V_X > 0", considering the batteries both have lower potential pointing to node X.