Two series-connected capacitors with capacities C1 = 2 µF and C2 = 4 µF
| May 1, 2021 | education
Two series-connected capacitors with capacities C1 = 2 µF and C2 = 4 µF are connected to a constant voltage source U = 120 V. Determine the voltage across each capacitor.
To begin with, we find the total capacitance, when the capacitors are connected in series, it is equal. 1 / Ctot = (1 / C1) + (1 / C2). From here we find Common.
Ctot = C1 * C2 / (C1 + C2) = 2 * 4 * 10 ^ -12 / (2 + 4) * 10 ^ -6 = 1.3 * 10 ^ -6 F.
Now you need to find the charge on the capacitors.
Q = Ctot * U = 1.3 * 10 ^ -6 * 120 = 156 * 10 ^ -6 Cl.
When capacitors are connected in series, their charges are the same.
Q = q1 = q2.
U1 = Q / C1 = 156 * 10 ^ -6 C / 4 * 10 ^ -6 = 39 B.
U2 = Q / C2 = 156 * 10 ^ -6 C / 2 * 10 ^ -6 = 78 B.
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