What is the maximum and minimum equivalent capacitance you can make using only three capacitors out of four?
You are given four capacitors, with capacitances 3000 pF, 0.0089 µF, 12000 pF, and 0.024 µF. Choosing from these four capacitors, what is the maximum and minimum equivalent capacitance that you can make using just three of these capacitors? Answer in Farads.
I know that the first one will be in a parallel circuit (maximum) and the second on will be in a series (minimum) but for some reason, I keep on getting the wrong answer after I convert all of them into Farads. Not sure what I'm doing wrong, please help?
- billrussell42Lv 78 years agoFavorite Answer
3000 pF, 0.0089 µF, 12000 pF, and 0.024 µF
or, switching to nF
3 nF, 8.9 nF, 12 nF, and 24 nF.
Max would be the 3 largest in parallel, which are the 24, 12 and 8.9, which add up to 44.9 nF, or 0.0000000449 F
Min would be the 3 smallest in series, which are the 3, 8.9, 12
1/c = (1/3) + (1/8.9) + (1/12) = 0.333 + 0.1124 + 0.0833
C = 1.89 nF
SI (metric) prefixes:
da (deca) = x10 (rarely used)
h (hecto) = x100 (rarely used)
k (kilo) = x1000 = e3
M (Mega) = x1000000 = e6
G (Giga) = x1000000000 = e9
T (Tera) = x1000000000000 = e12
P (Peta) = e15
E (Exa) = e18
Z (Zetta) = e21
Y (Yotta) = e24
d (deci) = /10 = e-1 (rarely used)
c (cent) = /100 = e-2 (rarely used except for cm)
m (milli) = /1000 = e-3
µ (micro) = /1000000 = e-6
n (nano) = /1000000000 = e-9
p (pico) = /1000000000000 = e-12
f (femto) = e-15
a (atto) = e-18
z (zepto) = e-21
y (yocto) = e-24