Three parallel plate capacitors C₁, C₂ and C₃ each of capacitance 5 μ F are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of C₁ capacitor is filled with a dielectric medium having dielectric constant of 4, is :

Solution

Physics sir · 4 steps

Fill C₁ with the dielectric first, then put C₁ and C₂ in series and add C₃ in parallel.

1. Read the circuit. C₁ and C₂ sit one after the other on the top branch, so they are in series. C₃ alone joins A to B on the bottom branch, so it is in parallel with that top pair.
2. Dielectric in C1. A dielectric of constant K that fills the whole gap multiplies the capacitance by K. So C₁ becomes 4 × 5 = 20 μF. C₂ and C₃ stay at 5 μF.

   C_(AB) = (C₁ C₂)/(C₁ + C₂) + C₃ = (20 × 5)/(20 + 5) + 5

3. Work it out. The series pair gives 100/25 = 4 μF. Adding C₃ in parallel gives 4 + 5 = 9 μF.
4. So. The effective capacitance between A and B is 9 μF, option (a).

The idea to keep. Update each capacitor for its dielectric first (C → KC when the gap is fully filled), then combine. In series the reciprocals add, and in parallel the capacitances add.

Answer from the NTA final answer key. Solution by PrepWiser.