Test: Combination of Capacitors - JEE 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
 

4μf and 6μf are in series (right side of AB) so 6x(4/6)+4=24/10μf
And now 5 μf and the resulting of above two are in parallel (as on different sides of AB) so 5+(24/10) =50+(24/10)=74/10=37/5μf

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=3C−−−(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