Test: Combination of Capacitors - NEET MCQ

Combination of Capacitors in Series
C(effective) = C/3 = 1

When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.

Mica capacitor. Mica capacitors have low resistive and inductive components associated with it. Hence, they have high Q factor and because of high Q factor their characteristics are mostly frequency independent, which allows this capacitor to work at high frequency.

C = Q/V
series connection splits the battery potential, hence....
V = V₁ + V₂
V₁ = Q/C₁ & V₂ = Q/C₂
Therefore, both have equal charge
 

For series combination,
1/C₁ = (1/C)+(1/C)+(1/C) [ Since all capacitors have equal capacitances]=3/C Or
C₁ = C/3−−−(i)
For parallel combination,
C₂ = C + C + C = 3 C−−−(ii)
Now,C₁ / C₂ = (C/3) / 3C = (C/3) × (1/3C) = 1/9

C1= 20×10µf

and C2= 30×10µf

in series Ceq = C1C2/(C1+C2)

Ceq = 20×10^(-6)×30×10^(-6)/20×10^(-6)+30^×10(-6)

Ceq= 12×10^(-6)f

As we know that Q = CV

Putting the values of C and V= 40V, we get

Q = (12 * 10^-6) * 40

= 480µC

Let C_p be the equivalent capacitance of parallel
C_p=C₁+C₂
16= C₁+C₂……..(1)
Let Cs be the equivalent of capacitance in series
1/C_s=(1/C₁) + (1/C₂)
Or, C_s=C₁C₂/C₁+C₂
Or,3=C₁C₂/16
Or,C₂C₁=48
C₂=48/C₁……..(ii)
16=C1+48/C1
(C₁²-16C₁+48)=0
(C₁-4)(C₁-12)=0
C₁=4 ; C₁=12
If,
C₁=4
C₂=12
If,
C₂=4
C₁=12
So, the answer is, Either 12μf or4μf

So, the equivalent circuit for the given circuit in the problem is as follows,

We know that the equivalent capacitance of a parallel combination set of the capacitors is the total sum of the individual capacitance.
So, C_P = 10 + 5 + 15 = 30 μF

Now this equivalent capacitance will be in series combination with the remaining one capacitor of magnitude 30μF.
So, the circuit will look like this,

So now the equivalent capacitance of the resultant circuit in between the point A and B is,

Now in the question it is given that the equivalent capacitance in between points A and B in the following combination is 5xμF.
So C_AB = 5x
⇒15 = 5x
⇒ 5x = 15
⇒ x = 15/5 = 3
⇒ x = 3

Hence the value of x is 3.