Question

two capacitor each of capacitance 4 m F are connected in series calculate their equivalent resistance​

Answer 1

Answer:

The required equivalent capacitance is 2 mF

Explanation:

Given :

Two capacitors each of capacitance 4 mF are connected in series.

To find :

the equivalent capacitance

Solution :

When the capacitors of capacitance C₁ , C₂ , C₃ ,.... are connected in series, the equivalent capacitance is given by

   $\sf \dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+....$

Here two capacitors of capacitance each 4 mF are connected in series.

Put C₁ = C₂ = 4 mF

   $\sf \dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2} \\\\ \sf \dfrac{1}{C_{eq}}=\dfrac{1}{4mF}+\dfrac{1}{4mF} \\\\ \sf \dfrac{1}{C_{eq}}=\dfrac{1+1}{4mF} \\\\ \sf \dfrac{1}{C_{eq}}=\dfrac{2}{4mF} \\\\ \sf \dfrac{1}{C_{eq}}=\dfrac{1}{2mF} \\\\ \sf C_{eq}=2 \ mF$

Therefore, the required equivalent capacitance is 2 mF

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Know more :

• When the capacitors of capacitance C₁ , C₂ , C₃ ,.... are connected in parallel, the equivalent capacitance is equal to the sum of all the individual capacitors.

       C = C₁ + C₂ + C₃ + ...

The voltage across all the capacitors is same.

The equivalent capacitance is more than any individual capacitance.

• When capacitors are connected in series,

The current flowing through each capacitor is same.

The equivalent capacitance is less than any individual capacitance connected in series.