Question

x)Two capacitors of 3mF and 6mF are connected in series. Their equivalent capacitance is?​

Answer 1

According to the given Question,

Two capacitors of magnitude 3mF and 6mF are connected in series.

Firstly, let us arrive at the generalized expression for series connection of capacitors.

Capacitance : Amount of charge stored in a device under a given potential.

$\implies \sf C = \dfrac{Q}{V} \\ \\ \implies \sf V = \dfrac{Q}{C}$

Now assume two capacitors are connected in series.

$\sf V_{net} = V_1 + V_2$

Now,

$\implies \sf \dfrac{Q}{C_{eq}} = \dfrac{Q}{C_{1}} + \dfrac{Q}{C_{2}}$

How does we know capacitors in series store same charge?

Since, current of same magnitude passes through any given connection in given time, Q can be conveniently amused to be same.

Therefore,

$\implies \sf \dfrac{1}{C_{eq}} = \dfrac{1}{C_{1}} + \dfrac{1}{C_{2}}$

For a series of 'n' capacitors,

$\implies \boxed{\boxed{\sf \dfrac{1}{C_{eq}} = \dfrac{1}{C_{1}} + \dfrac{1}{C_{2}} + \cdots + \dfrac{1}{C_{n}}}}$

Here, let the equivalent capacitance be C

$\longrightarrow \sf \: \dfrac{1}{C} = \dfrac{1}{3} + \dfrac{1}{6} \\ \\ \longrightarrow \sf \: \dfrac{1}{C} = \dfrac{1 + 2}{6} \\ \\ \longrightarrow \sf \: \dfrac{1}{C} = \dfrac{3}{6} \\ \\ \longrightarrow \boxed{ \boxed { \sf \: C = 2mF}}$

Answer 2

Answer:9uf



Explanation: