A capacitor with very large capacitance is in series with another capacitor with very small capacitance. What is the equivalent capacitance of the combination?

The equivalent capacitance of the combination is 1/C = 1/C1+1/C2 where C1 and C2 are the capacitance of both capacitors in series.

Explanation:

Let the capacitance of both capacitors be C1 and C2. For a series connected capacitors, same charge flows through the capacitors but different voltage flows through them.

Let Q be the amount of charge in each capacitors,

V be the voltage across each capacitors

C be the capacitance of the capacitor.

Using the formula Q = CV where V = Q/C... (1)

For the large capacitor with capacitance of the capacitor C1,

Q = C1V1; V1 = Q/C1...(2)

where V1 is the voltage across C1,

For the small capacitor with capacitance of the capacitor C2,

Q = C2V2; V2 = Q/C2 ... (3)

where V2 is the voltage across C2,

Total voltage V in the circuit will be;

V = V1+V2... (4)

Substituting equation 1, 2 and 3 in equation 4, we have;

Q/C = Q/C1 + Q/C2

Q/C = Q{1/C1+1/C2}

Since change Q is the same for both capacitors since they are in series, they will cancel out to finally have;

1/C = 1/C1+1/C2

This gives the equivalent capacitance of the combination.

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