Question

three capacitors of capacitance 2 uf connected in series. fine resultant capacitance​

Answer 1

Resultant capacitance of 3 capacitors in a series = 0.67 * 10^-6 F

Explanation:

When more than one capacitor is connected one after the another in a circuit, they are said to be connected in a series. To find the resultant capacitance of multiple capacitors connected in a series, we have to add the reciprocals of the individual capacitances, and then take the reciprocal of the sum found.

The unit for capacitance is Farad (symbol F), which is the equivalent of one coulomb per volt.

Capacitance of one capacitor = 2 μF

3 capacitors in series, so resultant capacitance = 1 / (1/2 + 1/2 + 1/2)  

 = 1 / (3/2)  

 = 2/3 μF

 = 2/3 * 10^-6 F

 = 0.67 * 10^-6 F

Answer 2

Answer:

Ceq = 0.67 uF

Explanation:

In general, the rule for combinational circuit of capacitors is that capacitances add up in parallel

Ceq = Ca + Cb

Now if they are in series circuit, then net reduces reduces.

Its formula is;

1/Ceq = 1/Ca + 1/Cb

In given case, we have three capacitors of 2uf which are connected in series. In this case, equivalent capacitance is;

1/Ceq = 1/2 + 1/2 + 1/2

1/Ceq = 3/2

Ceq = 0.67 uF