Question

The equivalent capacitance of two capacitors in series is 3 microF and in parallel is 16 microF. Their individual capacitance are ………microF and …….microF.​

Answer 1

Answer:

Given that,

The equivalent capacity of two capacitor

in series is 3 uF and in parallel is 16 u F

To find,

Their individual capacities.

Solution,

For series combination, equivalent

capacitance is given by:

1

Cs

C1 C

1

We have, C, = 3 puF

So,

3 CT (1)

For parallel combination, the equivalent

capacitance is given by:

Cp = C1 + C2

We have, Cp = 16 puF

So,

So,

16 = C1 + Ca ..(2)

We need to solve equation (1) and (2) to

find the value of C and C2. If we solve

them, we get

C=4 pF

C2 = 12 juF

So, their individual capacities are 4 and

12 microfarads.

Answer 2

Given:

The equivalent capacitance of two capacitors in series is 3 micro-F and in parallel is 16 micro-F.

To find:

Individual Capacitance?

Calculation:

Let the Capacitance be C_(1) and C_(2).

Now, we know that :

• For series : $\dfrac{C_{1}C_{2}}{C_{1} + C_{2}} = 3$

• For parallel : $C_{1}+C_{2} = 16$

So, putting values in 1st equation:

$\dfrac{C_{1}C_{2}}{C_{1} + C_{2}} = 3$

$\implies \dfrac{C_{1}C_{2}}{16} = 3$

$\implies C_{1}C_{2} = 48$

Now, putting the value of C_(1) or C_(2), we get:

${(C_{1})}^{2} - 16C_{1} + 48 = 0$

$\implies {(C_{1})}^{2} - (12 + 4)C_{1} + 48 = 0$

$\implies \: (C_{1} - 12)(C_{1} + 4) = 0$

So, C_(1) = 12 uF and C_(2) = 16-12= 4 uF.