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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

The effective capacitances of two capacitors are 3 \muF3 \text{ \mu F} and 16 \muF16 \text{ \mu F}, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

A

10 \mu F, 6 \mu F

B

8 \mu F, 8 \mu F

C

12 \mu F, 4 \mu F

D

1.2 \mu F, 1.8 \mu F

Step-by-Step Solution

  1. Formulas:
  • Parallel combination: Cp=C1+C2C_p = C_1 + C_2
  • Series combination: Cs=C1C2C1+C2C_s = \frac{C_1 C_2}{C_1 + C_2}
  1. Given Data:
  • Cp=16 \muFC_p = 16 \text{ \mu F} (Sum of capacitances)
  • Cs=3 \muFC_s = 3 \text{ \mu F}
  1. Set up Equations:
  • From parallel: C1+C2=16C_1 + C_2 = 16
  • From series: C1C216=3  ⟹  C1C2=48\frac{C_1 C_2}{16} = 3 \implies C_1 C_2 = 48
  1. Solve: We need two numbers that add up to 16 and multiply to 48. Using the quadratic equation x2−(sum)x+(product)=0x^2 - (sum)x + (product) = 0: x2−16x+48=0x^2 - 16x + 48 = 0 (x−12)(x−4)=0(x - 12)(x - 4) = 0 Therefore, x=12x = 12 or x=4x = 4.
  2. Verification:
  • Sum: 12+4=1612 + 4 = 16 (Matches CpC_p)
  • Series: 12×412+4=4816=3\frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 (Matches CsC_s)

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEeffectivecapacitancescapacitorsconnectedseries

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