Capacitor equation

[jimmyjazz]

Can anyone help with this college equation?
3 x capacitors in parallel - 10pF (To the power of -12), 15 uF (-6), 25 nF (-9). What does Ct = ? Tutor says 10 (-6), I think 10 (-14).
Thanks

 

[Geoff]

Hello Jimmyjazz.
I get 9.99pF (x10-12) Though I could be wrong! I often am

 

[jimmyjazz]

Thanks Geoff. If I put the whole equation into calc. I get 10x-14. If I add up top & bottom of equation separately 10-18/10-6 = x10-12. Reckon this is right.

 

[jimmyjazz]

.....just remembered, isn't Ct ment to be less than lowest individual resistance?

 

[bomjac]

Capacitors in series - just add them all together.
Capacitors in parallel - 1/[(1/a)+(1/b)+(1/c)......]

Would make your total to be ~ 15uF.

Or more exact 15.025010 uF

 

[Geoff]

Yeah! indeed it is. I always use the (1 over button) on my calc for each value, And again after Equal. Flippin Maths!!!!

 

[Knobhead]

I get 10.004pF

Think it's time we all went back to school!

 

[JUD]

Capacitors in series - just add them all together.
Capacitors in parallel - 1/[(1/a)+(1/b)+(1/c)......]

Would make your total to be ~ 15uF.

Or more exact 15.025010 uF

That should be the other way around.

Capacitors in series = 1 ÷ ((1 ÷ a) + (1 ÷ b) + (1 ÷ c)......)
Capacitors in parallel = Add them all together.

Although your answer is correct.

[10 x 10^(-12)] + [15 x 10^(-6)] + [25 x 10^(-9)] = 15.02501uF

 

[bomjac]

Whoops....was thinking the right way, calculating the right way and write it wrong lol.

 

[jimmyjazz]

Thanks to all, its bloody difficult enough without the tutor showing you the wrong way to do it!