Question 64 of 180

Three capacitances, each of 3 $\mu \text{F}$, are provided. These cannot be combined to provide the resultant capacitance of :

A$2\mu \text{F}$B$6\mu \text{F}$C$1.5\mu \text{F}$D$1\mu \text{F}$

Show answer and solution▾

Answer: (B) $6\mu \text{F}$

Equivalent capacitance will be maximum for parallel combination and Minimum for series combination

$∴$ Possibilities are

1) $\text{C}=1\mu \text{F}$

2) $\text{C}=9\mu \text{F}$

3) $\dfrac{1}{\text{C}}=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{3}{6}$

$∴\text{C}=2\mu \text{F}$

4) ​ $\text{C}=\dfrac{\text{C}}{2}+\text{C}=\dfrac{3}{2}\times \text{C}$

$=\dfrac{3}{2}=\text{1.5}\mu \text{F}$

JEE Main Physics practice question on Capacitors (Grouping of capacitors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.