Answer:
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.
