Question

Five identical capacitors having which capacitance 5F are connected in series find the effective capacitance​

Answer 1

Given :

Five identical capacitors having which capacitance 5F are connected in series

To be found:

The effective capacitance​

⇒ Capacitors in series :

For n capacitors connected in series the equivalent capacitance

$\frac{1}{c_{eq} } = \frac{1}{c_{1} } + \frac{1}{c_{2} } + ..... + \frac{1}{c_{n} }$

⇒ Capacitors in Parallel :

For n capacitors connected in parallel, the equivalent capacitance is given by

$c_{eq} = c_{1} + c_{2} + .... c_{n}$

Solution :

We have five capacitors, of having capacitance 5F.

When they are connected in series

$\frac{1}{c_{eq}}=\frac{1}{c_1}+\frac{1}{c_2}+\frac{1}{c_3}+\frac{1}{c_4}+\frac{1}{c_5}$

Since all capacitors have same capacitance 5F

It means

$\sf\:C_1 =C_2 =C_3 =C_4=C_5 =5F$

$\implies \frac{1}{c_{eq}}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$

$\implies \frac{1}{c_{eq}}=\frac{1+1+1+1+1}{5}$

$\implies \frac{1}{c_{eq}}=\frac{5}{5}$

$\boxed{\sf C_{eq} = 1F}$

Therefore,

Effective capacitance is 1 F

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Extra information

1) Series combination

In Series combination, each capacitor has an equal charge for any value of capacitances.

2) Parallel Combination

When capacitors are connected in parallel, the potential difference across each capacitor is the same.

Answer 2

Given ,

The five identical capacitors having which capacitance 5 F are connected in series

We know that , the equivalent capacitance in series combination is given by

$\boxed{ \tt{\frac{1}{C} = \frac{1}{C'} + \frac{1}{C''} + ...}}$

Thus ,

1/C = 1/5 + 1/5 + 1/5 +1/5 + 1/5

1/C = (1 + 1 + 1 + 1 + 1)/5

1/C = 5/5

C = 1 farad

Hence , the equivalent capacitance is 1 farad