Question

What is the equivalent capacitance of the three capacitors in the figure (figure 1)?

Answer 1

The equivalent capacitance of the combination is  $\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}$  where C1 and C2 are the capacitance of both capacitors in series.

What is equivalent capacitor?

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,

$Q = C_1V_1;$

$V_1 = \dfrac{Q}{C_1}...(2)$

where $V_1$ is the voltage across $C_1,$

For the small capacitor with capacitance of the capacitor $C_2$,

$Q = C_2V_2$;

$V_2 = \dfrac{Q}{C_2} ... (3)$

where $V_2$ is the voltage across $C_2$,

Total voltage V in the circuit will be;

$V = V_1+V_2... (4)$

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

$\dfrac{Q}{C} = \dfrac{Q}{C_1} +\dfrac{ Q}{C_2}$

$\dfrac{Q}{C} = Q({\dfrac{1}{C_1}+\dfrac{1}{C_2})$

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

$\dfrac{1}{C} = ({\dfrac{1}{C_1}+\dfrac{1}{C_2})$

This gives the equivalent capacitance of the combination.

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