A series circuit consists of a 0.046 µF capacitor, a 0.140 µF capacitor, and a 380 V battery.?
Find the charge for the following situations.
(a) on each of the capacitors
____ µC (0.046 µF capacitor)
____ µC (0.140 µF capacitor)
(b) on each of the capacitors if they are reconnected in parallel across the battery
____ µC (0.046 µF capacitor)
____ µC (0.140 µF capacitor)
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All three elements are in series ?
The equivalent C (in nF) = 46*140/(46+140) = 35 nF
Q = CV = 35 nF x 380v = 13200 nC pr 13.2 µC
same charge is on both caps.
In parallel
Q = CV = 380 x 46 nF = 17.4 µC
Q = CV = 380 x 140 nF = 53 µC
a) when in series, V across C1= 380x(0.140/(0.046+0.140)= 286.02 V
Charge across C1 = 0.046 x 10^-6 x 286.02= 13.157 µC
V across C2= 380 - 286.02= 93.98 V
Charge across C2 = 0.140 x 10^-6 x 93.98= 13.157 µC
b) when in parallel, Charge across C1 = C1 x V= 0.046 x 10^-6 x 380 = 17.48 µC
Charge across C2 = C2 x V= 0.140 x 10^-6 x 380 = 53.2 µC
w = 2pi*f = 2pi*60 = 377 rad/s Z = R + j(wL - a million/wC) the area j = sqrt(-a million) (in case you do no longer understand this line, pass directly to the subsequent) |Z| = sqrt[R^2 + (wL - a million/wC)^2] |Z| = sqrt{4.84e4 + [377*27e-6 - a million/(377*16e-6)]^2} = sqrt[4.84e4 + (a million.02e-2 + a million.66e2)^2] = 275.6 ohms (be conscious thet the inductor seems pretty much like only quite twine with L=0) Then rms i = a hundred and twenty/275.6 = 0.435A answer