Solution Exercises Capacitor

Solution 2:

Wrong way: X_C = 1/ (2πfC), This means that each of the capacitors would have an infinitely high resistance. Calculating the voltage drop according to Ohm's law does not lead to the desired result.

Right way: Q = C ⋅ V   or   V = Q / C

Determination of charge quantity Q

Q = C_total ⋅ V = 3,38 µF ⋅ 10V = 33,8 µC

.  1     =    1  +   1   +   1   ⇒   C_total  = 3,38 µF
C_total        C₁       C₂      C₃

The charge is the same on all capacitors. This gives:

V₁  =  Q  =   33,8 µAs  = 5,63 V
.         C₁       6 µAs/V

V₂  =  Q  =   33,8 µAs  = 2,82 V
.         C₂       12 µAs/V

V₃  =  Q  =   33,8 µAs  = 1,54 V
.         C₃       22 µAs/V