Question

the capacities of three capacitors are in ratio of 1:2:3 .their equivalent capacity in parallel is greater than their equivalent capacity in series by 60/11 pf. calculate their individual capacitance

Answer 1

parallel capacitance Cp = C+2C+ 3C = 6C
series capacitance 1/Cs = 1/C + 1/2C+1/3C
= 11/6C
therefore Cs = 6C/11
from question Cp = 60/11 +Cs
therefore. 6C = 60/11+6C/11
so. C = 1
therefore actual capacitance 1,2 and 3 pf

Answer 2

Concept:

The ratio of the quantity of electric charge stored on a conductor to the difference in electric potential is known as capacitance.

Given:

Three capacitors in the ratio $1:2:3$

To Find:

Individual Capacitance of the capacitors

Solution:

Let the capacitance be $C, C_{1} , C_{2}$ = 1:2:3

The values will be $x, 2x, 3x$

when arranged in series,

$C_{s} = [(1 / x)+(1 / 2 x)+(1 / 3 x)]^{-1}=[(6+3+2) /(6 x)]^{-1}$

When arranged in parallel,

$C_{p}=x+2 x+3 x=6 x$

Since,

$C_{p} = \frac{60}{11} + C_{s}$

$6 x-(6 x / 11)=(60 / 11) \\ {[(60 x) /(11)]=(60 / 11) } \\ x=1$

The individual capacitance of the three capacitors is $1 \mu \mathrm{F}, 2 \mu \mathrm{F}, 3 \mu \mathrm{F}$ .

                                                               

                                                                                                          #SPJ2