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The equivalent capacitance of the combination is [tex]\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex] where C1 and C2 are the capacitance of both capacitors in series.
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,
[tex]Q = C_1V_1;[/tex]
[tex]V_1 = \dfrac{Q}{C_1}...(2)[/tex]
where [tex]V_1[/tex] is the voltage across [tex]C_1,[/tex]
For the small capacitor with capacitance of the capacitor [tex]C_2[/tex],
[tex]Q = C_2V_2[/tex];
[tex]V_2 = \dfrac{Q}{C_2} ... (3)[/tex]
where [tex]V_2[/tex] is the voltage across [tex]C_2[/tex],
Total voltage V in the circuit will be;
[tex]V = V_1+V_2... (4)[/tex]
Substituting equation 1, 2 and 3 in equation 4, we have;
[tex]\dfrac{Q}{C} = \dfrac{Q}{C_1} +\dfrac{ Q}{C_2}[/tex]
[tex]\dfrac{Q}{C} = Q({\dfrac{1}{C_1}+\dfrac{1}{C_2})[/tex]
Since change Q is the same for both capacitors since they are in series, they will cancel out to finally have;
[tex]\dfrac{1}{C} = ({\dfrac{1}{C_1}+\dfrac{1}{C_2})[/tex]
This gives the equivalent capacitance of the combination.
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