Question

Ans. (1) lut (1) yur
21. Two capacitors have an equivalent capacitance of 24 uF when joined in parallel and of 6 F when
joined in series. What are their individual capacitances ?
Ans. 12 pF, 12 uF
22. We have two variable capacitors: each of them can be varied from 10 pF to 500 pF What minimum
and maximum capacitances can be obtained by combining them?
Ans. 5 pF 1000 pF​

Answer 1

Answer:

12 & 12

5 & 1000

Explanation:

Let say individual capacitance =

C₁  & C₂

Joined in Parallel  = C₁  + C₂ = 24

joined in series = 1/(1/C₁  + 1/C₂)  = 6

=> C₁C₂/(C₁  + C₂)  = 6

=> C₁C₂ = 6(C₁  + C₂)

=> C₁C₂ = 6 * 24

=> C₁C₂ = 144

=> C₁ (24 - C₁ ) = 144

=> 24C₁  -C₁ ² = 144

=> C₁ ²  - 24C + 144 = 0

=> C₁  - 12C - 12C + 144 = 0

=> (C₁ - 12)(C₁ - 12) = 0

=> C₁ = 12

C₁  + C₂ = 24  => C₂ = 12

individual capacitance = 12

10 pF to 500 pF

C₁  & C₂

in  Parallel = C₁  + C₂

this will be max when both C₁  & C₂ will be max

=>  C₁  = C₂ = 500pF

=> C₁  + C₂ = 1000 pF

in  Series =  C₁ *C₂ /(C₁  + C₂ )

Will be minumum when  C₁  = C₂

= C₁/2 = C₂/2

= 10/2 = 5 pF