CAPACITANCE_JEE-Main 2020_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2020

02-09-2020 S2

Question

A $10 \mu \mathrm{~F}$ capacitor is fully charged to a potential difference of 50 V . After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V . The capacitance of the second capacitor is

Select the correct option:

$20 \mu \mathrm{~F}$

$10 \mu \mathrm{~F}$

$15 \mu \mathrm{~F}$

$30 \mu \mathrm{~F}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & V_{\text {final }}=\frac{C_1 V}{C_1+C_2} \\ & C_2=15 \mu F\end{aligned}$

Question Tags

JEE Main

Physics