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Hell-o, New guy here,
just to jump in on the capacitor problem
2 capacitors in series of 1MFD and 2MFD
Scott is correct.
Ct= .666 MFD
Qt = Q1 = Q2
then Qt = Ct x E (assume E=12 VDC)
.667E-6 x 12 = 8E-6

V1 = Qt/C1 = 8E-6/1E-6 = 8 volts
V2 = Qt/C2 = 8E-6/2E-6 = 4 volts

E = V1 + V2 thus 8 + 4 = 12
and thats the way I see it.
Tom

Welcome to ECN Tom.

Yes, that's the answer. The charge, Q, in coulombs is equal to the product of capacitance and voltage.

The capacitors are in series, therefore each will acquire the same charge. Thus the 1uF capacitor must end up with twice the voltage of the 2uF cap.