## Wednesday, August 12, 2009

### Capacitance - taking things on a step further

Countdown Timer: 7 weeks 6 days to go

You remember we touched on capacitors, and storing charge, and bin lorries? That was just over two weeks ago - read it here if you need to refresh your memory. Well now we're going to take things on a step.

I learned a very valuable lesson in Brian's QTH on Monday night which has helped me solve some capacitance questions quite easily. Here it is, so read carefully and take it in:

With resistors, when in series add them up, when in parallel, its R1xR2/R1+R2. Got that? Basic Ohm's law stuff.

Well, with capacitors, it's the OPPOSITE!

When in series, its C1xC2/C1+C2

And when in parallel, you add their values.

This basic knowledge helped me answer another question on the Sample Paper, which I present below: We see that capacitors C2 and C3 are in series, so we need to work out their overall capacitance by using: C2xC3 / C2+C3 = 20x20/20+20 = 400/40 = 10. So 10 nF is the overall capacitance of C2 and C3. They are in parallel with C1, so we just add 10 to the value of C1, which is 5 nF, giving 15 nF.

So B - 15 nF - is the correct answer.

I now have in my possession another book, called "towards the Radio Amateurs' Examination" by John Bowyer, published by Stam Press Limited. I tried the 10 questions on capacitance and excelled myself by getting 9 out of 10 or 90%. I will give some examples here.

1. The parallel plates of a capacitor are separated by 2mm of air. If this separation is increased to 4mm, the new value of capacitance will be:

a four times the original capacitance
b double the original capacitance
c one-half the original capacitance
d one-quarter the original capacitance

I knew, having read it, that the capacitance reduces as the distance between the plates increases, but this is actually expressed in the formula:

Capacitance of a capacitor = permittivity x plate area / distance between plates

otherwise expressed as 1 / distance between plates

Therefore, if the distance is doubled, then C must be halved. So the answer is (c).

2. Two capacitors, of value 10 µF and 40 µF respectively, are connected in series. An equivalent capacitor will have a value of:

a 8 µF
b 30 µF
c 50 µF
d 400 µF

Remember that with capacitors it's the opposite to resistors, so when capacitors are in series, we use C1xC2/C1+C2, so 10x40/10+40 = 400/50 = 8.

The answer is (a) 8 µF !