Dibujar un circuito con dos condensadores de 8 y 5 nf (nanofaradios), respectivamente en paralelo, con una fuente de 20voltios, hallar la
capacitancia equivalente, con su circuito, hallar las cargas en cada uno de
los capacitores, la caída de tensión en cada capacitor (demostración).

CT = 470nF + 1000nF = 1470nF or 1.47μF

Explanation:

When capacitors are connected together in parallel the total or equivalent capacitance, CT in the circuit is equal to the sum of all the individual capacitors added together. This is because the top plate of capacitor, C1 is connected to the top plate of C2 which is connected to the top plate of C3 and so on.  The same is also true of the capacitors bottom plates. Then it is the same as if the three sets of plates were touching each other and equal to one large single plate thereby increasing the effective plate area in m2.

Since capacitance, C is related to plate area ( C = ε(A/d) ) the capacitance value of the combination will also increase. Then the total capacitance value of the capacitors connected together in parallel is actually calculated by adding the plate area together. In other words, the total capacitance is equal to the sum of all the individual capacitance’s in parallel. You may have noticed that the total capacitance of parallel capacitors is found in the same way as the total resistance of series resistors.

The currents flowing through each capacitor and as we saw in the previous tutorial are related to the voltage.