Test: Combination of Capacitors (NCERT) - NEET MCQ

To get equivalent capacitance 6 μF. Out of the 4 μF capacitance, two are connected in series and third one is connected in parallel.

Correct answer is A.

Form the figure C₁, C₂, C₃ are connected in series.


Now C_s is in parallel with C₄
∴ equivalent capacitance C_eq = C_s + C₄

Charge on each capacitor C₁, C₂, C₃ is same = q
Let charge on C₄ be q.


Also qC₄ = 600
or q = C₄ × 600μC = 20 × 600μC = 12 × 10⁻³C

Since all the capacitors are connected in series,
∴ equivalent capacitance is

Charge on each capacitor is same  (∵ V = 10V)
q = C_eqV = 1 × 10 = 10μC

Minimum number of capacitors in each row = 1000/250 = 4
Therefore, 4 capacitors connected in series.
If there are m such rows, that total capacity
= m × 2 = 16 ∴ m = 16/28
∴ minimum number of capacitors = 4 × 8 = 32

When a charged capacitor of capacitance C₁  is connected in parallel to an uncharged capacitor of capacitance C₂, the charge is shared between them till both attain the common potential which is given by, common Potential = q₁ + q₂/C₁ + C₂ = q₁ + q₂/C₁ + C₂ = C₁V/C₁ + C₂

C_p = 2 + 4 = 6μF

Total Q = CV = 3 × 12 = 36μF
Voltage across 6μF capacitor = 36μC/6μF = 6V
∴ Voltage across each of 2μF and 4μF capacitors = 12V - 6V
V = 6V

 

When the two capacitors charged to same potential are connected in series, then total potential difference V′ = V₁ + V₂ = V + V = 2V

From the figure, when K₁ ​is closed and K₂ is open, C₁ ​is charged to potential V acquiring a total charge q = C₁V When K₁ is open and K₂ ​ is closed, battery is cut off C₁ ​and C₂ ​are in parallel. The charge on C₁ is shared between C₁ ​and C₂ ​such that V₁ = V₂ ​
As there is no loss of charge q₁′ ​+ q₂′ ​= q