Question

30. Two 50 µF capacitors are connected in series. The equivalent capacitance is;

Answer 1

Answer:

The required equivalent capacitance is 25 µF

Explanation:

Given :

Two 50 µF capacitors are connected in series.

To find :

the equivalent capacitance

Solution :

When capacitors of capacitances C₁, C₂, C₃, . . . Cₙ are connected in series, the equivalent resistance (C) is given as

 $\sf \dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+...+\dfrac{1}{C_n}$

Let C₁ = C₂ = 50 µF

Then, the equivalent resistance is

 $\sf \dfrac{1}{C}=\dfrac{1}{50}+\dfrac{1}{50} \\\\ \sf \dfrac{1}{C}=\dfrac{1+1}{50} \\\\ \sf \dfrac{1}{C}=\dfrac{2}{50} \\\\ \sf \dfrac{1}{C}=\dfrac{1}{25} \\\\ \longrightarrow \sf C=25 \ \mu F$

Therefore, the equivalent resistance is 25 µF