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Refer to the circuit below. Where C

1

​

=1.00×10

−6

FC

2

​

=2.00×10

−6

FC

3

​

=3.00×10

−6

C

4

​

=4.00× 10

−6

F a. What is the equivalent Capacitance of the system? b. What is the potential difference across on C

2

​

?

Answer :

The equivalent capacitance of the system is [tex]4.81*10^{-7} F[/tex] and the pd across C2 is 4.15 times the total pd in the circuit

a. The equivalent capacitance of the system can be found by adding up the individual capacitances in series or parallel. In this case, the capacitors C1, C2, C3, and C4 are connected in series. Formula for equivalent capicitance is:


1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4

Substituting the given values:

[tex]1/Ceq = 1/(1.00*10^{-6}) + 1/(2.00*10^{-6}) + 1/(3.00*10^{-6}) + 1/(4.00*10^{-6})[/tex]

Simplifying the equation:

[tex]1/Ceq = 10^6/(1.00 + 0.50 + 0.33 + 0.25) \\

1/Ceq = 10^6/(2.08) \\

Ceq = 4.81×10^{-7} F
[/tex]


b. Formula for potential difference across the capacitor C2 is:
Vc2 = V × (C2/Ceq)

Where Vc2 is the potential difference across C2, V is the total potential difference in the circuit, C2 is the capacitance of C2, and Ceq is the equivalent capacitance of the system.

Substituting the given values:

[tex]Vc2 = V * (2.00*10^{-6} F / 4.81*10^{-7} F)[/tex]

Simplifying the equation:

Vc2 = V × 4.15

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