Consider the combination of 2 capacitors C₁ and C₂ with C₂ > C₁, when connected in parallel, the equivalent capacitance is 15/4 times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, C₂/C₁.

Solution

Physics sir · 4 steps

Parallel over series can never drop below 4, so a ratio of 15/4 is impossible for any two capacitors.

1. Write both combinations. In parallel, Cₚ = C₁ + C₂. In series, Cₛ = (C₁ C₂)/(C₁ + C₂). Dividing the first by the second gives Cₚ/Cₛ = (C₁ + C₂)²/(C₁ C₂).
2. Use the ratio. Put r = C₂/C₁ and divide the top and bottom by C₁². The condition becomes (1 + r)²/r = 15/4, which rearranges to a quadratic in r.

   4r² - 7r + 4 = 0, D = (-7)² - 4 × 4 × 4 = 49 - 64 = -15

3. Why it fails. The discriminant is negative, so no real r works. You can also see why: (1 + r)² - 4r = (1 - r)² ≥ 0, so (1 + r)²/r ≥ 4 for every positive r. The smallest value is exactly 4, when C₁ = C₂, and 15/4 = 3.75 sits below it.
4. So. No pair of real capacitors gives this ratio. The answer is (b), no solutions.

The idea to keep. For two capacitors, Cₚ/Cₛ = (C₁ + C₂)²/(C₁ C₂) is always at least 4, and equals 4 only when they are equal. Check this lower bound before you start solving.

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