- ?Lv 71 month ago
Total capacitance is 1/(1/C1 + 1/C2 + 1/C3) = 0.6242 pF.
The charge stored on each capacitor is
CV = (0.6242 pF)(1700 V) = 1.061 nC.
The voltage associated with each capacitor can now be found
by dividing 1.061 nC by the individual capacitance values.
For instance, V1 = (1.061 nC)/(3.15 pF) = 337 V.
You can get V2 and V3 for yourself; the total of V1 + V2 + V3 will be 1700 V.
That's assuming that what I'm reading as 1700 was not 700 with a random slash in front of it. Hard to tell.
If you wonder why each capacitor stores the same quantity of charge, think about the single piece of metal between C1 and C2. The piece of metal is neutral, but some of the electrons are driven to one end. The positive charge on one end must equal the negative charge on the other end.
- billrussell42Lv 71 month ago
can't read the values of the caps. my guess is 3, 2, 1 pF with 1000 volts
caps in series add via:
1/C = 1/C1 + 1/C2 + 1/C3
1/C = 1/3 + 1/2 + 1/1 = 0.33+0.5+1 = 1.83
C = 0.54 pF
I can read that you want total C, V, Q
total C is above
total V is 1000 volts
total Q = CV = 0.54 x 1000 = 0.54 nC
if you want the values for each, which is not asked for...
Q is same on all three
V1 = Q/C1 = 0.54 nC / 3 pF = 180 volts
V2 = Q/C2 = 0.54 nC / 2 pF = 270 volts
V3 = Q/C3 = 0.54 nC / 1 pF = 540 volts
which adds up to 1000, with some rounding error