Competition - post a photo you've taken of nature >>

capacitance

Scroll to see replies

So in series, all capacitors have the same charge, but in parallel, the capacitors have different charges. How will I work this out for the first part of the second question? Q=CV using the individual capacitance and the voltage as 100V for each?

100V is used for all capacitors in the parallel case only. For series capacitors, you need to calculate the p.d. for each.

Go back and read my answers again. All the information you need is there.

Parallel example

Parallel capacitors are free to take up the maximum charge governed by the supply voltage. Each capacitor can store a different level of charge. V is the common variable for all of them. Q is a function of (dictated by) V & C.

Any component placed in parallel with the supply must have the supply voltage placed across it. The source of their current and the place to which that current flows are one and the same. i.e. in this case, the supply is 100V so all parallel capacitors see 100V.

The same applies for any component in parallel: the voltage across the nodes of parallel components is identical.

Think of it like the current analogy to water pressure where the current entering a junction splits and part flows down one pipe while the rest flows down the other.
The individual parallel components resist the flow and hence the currents flowing down each path are determined by the individual components (be they resistors, capacitors, indictors, semiconductors et al).

The total current entering the junction (node) must be the sum of the currents flowing down each separate path. This is Kirchoff's Current Law (KCL)

When the paths recombine, the current flowing out of the node joining the parallel paths must be the same as the sum of the currents in the individual parallel paths. Again this is KCL.

The water analogy is where the term electric current historically came from in the first place.

Series example

As per my previous description, the smallest capacitor limits the total charge the series capacitors can store.
Because of this, Q is the common variable and hence V is a function of Q & C.

a) First work out the total effective capacitance in the series path.

b) Next calculate the total charge held by the effective capacitance by using that value and the p.d. across the series combination. In this case 100V

c) Calculate the p.d. developed across each individual capacitor by using the total charge value and each individual capacitor value.

d) Total energy is once again the effective capacitance value and the voltage across all parallel capacitors. (CV²/2 or QV/2

e) Individual stored energy is then calculated by subbing the individual p.d.'s calculated in c) above and the individual capacitor values.

(edited 9 years ago)

Reply 21

rm2

I have already calculated the voltage in series. I was trying to find the charge for the capacitors in parallel and got:

Q = CV so using the capacitance and the voltage (100V for all because they're in parallel)

Q1 = 3×10^-6 × 100 = 3×10-⁴ C

Q2 = 6×10-⁴ C

Q3 = 12×10-⁴ C

capacitance

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you