Charge on Capacitor in Pure Capacitance Circuit?

In summary: This results in the final answer being slightly higher than what is required. In summary, the homework statement asks for the charge on one of eight capacitors when a 72 volt source is attached. The Attempt at a Solution states that the top two pairs of capacitors are in parallel, so if we simplify those, then we end up with three capacitors in series, one of which is 8.9 microFarads, and the other two that are 17.8 microFarads. The charge on all three capacitors is going to be the same (since they're in series) and so if I combine all three of these capacitors in series, I get 4.45 microFarads equivalent capacitange. If I
  • #1
eprparadox
138
2

Homework Statement


I apologize for having to upload an attachment but I couldn't describe this circuit. Basically, there is set of capacitors and the question is asking for the charge on one of them, when a 72 volt source is attached. All capacitances are 8.9 microFarads.


Homework Equations


Q = C*V


The Attempt at a Solution


I believe the top two pairs of capacitors are in parallel. So if we simplify those, then we end up with 3 capacitors in series, one of which is 8.9 microF, and the other two that are 17.8 microF.

Then from there, I can say the charge on all 3 capacitors is going to be the same (since they're in series) and so if I combine all 3 of these capacitors in series, I get 4.45 microF equivalent capacitange.

And if I have a 72 volt source, then I get Q = Ceq*V = 320 microC. So then the voltage across the parallel combination (which is 17.9 microF) becomes V = 320 microC / 17.9 microF = 17.9 V. This is the same voltage that is across the original C2 capacitor so the final charge on C2 should work out to be Q = C * V = 8.9 microF * 17.9 V = 159 microCoulombs as my Final answer.

I hope that makes sense. I just want to know if my mindset is correct in working through this.

Any help from you guys and girls is greatly appreciated.
 

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  • #2
This is entirely correct. I would do all the computations with C and V, only substituting numbers for the final answer.
 
  • #3
There is also a simpler way of approaching this problem - I was tempted to post the whole method, but worried about getting in trouble for giving a direct answer.

Suffice it to say that four identical capacitors connected as two groups in series, each of two capacitors in parallel equals...what exactly?

This should lead rapidly on to finding what fraction of the supply voltage falls across C2. It's a simple fraction, giving a voltage very close to your value of 17.9V.

I would agree with willem2 that you should try to leave calculations to the end. Although your working is basically correct there seems to be a very small arithmetic error.
 

1. What is a pure capacitance circuit?

A pure capacitance circuit is a type of electrical circuit that consists only of capacitors and does not contain any resistors or inductors. The circuit behaves as a purely reactive element, storing and releasing electrical energy in the form of an electric field.

2. How does a pure capacitance circuit work?

In a pure capacitance circuit, the capacitor stores electrical energy in the form of an electric field between its plates. When a voltage is applied to the circuit, the capacitor charges up and stores this energy. When the voltage is removed, the capacitor discharges and releases the stored energy.

3. What is the impedance of a pure capacitance circuit?

The impedance of a pure capacitance circuit is purely imaginary and is given by the formula Z = 1/jωC, where j is the imaginary unit, ω is the frequency of the applied voltage, and C is the capacitance of the circuit. This means that the impedance is solely dependent on the frequency and capacitance, and there is no resistance component.

4. How does the frequency affect a pure capacitance circuit?

The frequency of the applied voltage has a direct effect on the behavior of a pure capacitance circuit. As the frequency increases, the reactance of the circuit decreases, allowing more current to flow through it. At lower frequencies, the reactance is higher, and less current can flow.

5. What are the practical applications of pure capacitance circuits?

Pure capacitance circuits have many practical applications in electronics, including power factor correction, filtering and smoothing circuits, and timing circuits. They are also used in touch sensors, capacitive sensors, and as energy storage devices in electronic devices such as cameras and flashlights.

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