What is the capacitance in nF?
Capacitance is charge/volts, so 27 x 10^-6 / 450 V = 6x10^-8
and then multiply that answer by 10^-9 to get nF?
2007-10-19 06:24:13 · 4 answers · asked by prime 3 in Science & Mathematics ➔ Engineering
Correct but not multiply by 10^-9 but divide. So that you would get 60 nF.
You can look this way: 6*10^-8 = 60*10^-9 F but 10^-9 is nF, so 60*10^-9 F = 60 nF
2007-10-19 06:28:37 · answer #1 · answered by Alexey V 5 · 1⤊ 1⤋
from Farads to nF
yes, divide by 10^9
you got a wrong answer (from Alexy V)
divide by 10^9 OR multiply by 10^-9
both statements are correct
2007-10-19 09:59:01 · answer #2 · answered by Anonymous · 0⤊ 2⤋
Q=C*V so 27*10^-6 /450 = 27/0.450 *10^-9 F=60nF
2007-10-19 06:31:25 · answer #3 · answered by santmann2002 7 · 0⤊ 2⤋
a) ability C = eo A / d = q / V, so as that as quickly as 'd' is doubled, C gets halved, then for preserving q consistent V must be halved, for this reason V turns into 2 hundred volt. b) by using an analogous relation, hence additionally q must be halved.
2016-12-15 03:58:13 · answer #4 · answered by ? 4 · 0⤊ 0⤋