Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination.

Answers (1)

The energy stored in capacitor, when connected in series combination;

We have given, $C_{S}=6{P}F=6\times 10^{-12}F$

Hence, the energy stored $(E_{S})$ in capacitor ;-

$E_{S}=\frac{1}{2}C_{S}V^{2}$

$=\frac{1}{2}\times 6\times 10^{-12}\times 50\times 50$  (Given; V=50)

$=7500\times 10^{-12}J$

$E_{S}=7.5\times 10^{-9}J$

Now, when capacitors are connected in parallel,

the value of the capacitor is given as,

$C_{P}=-24{P}F=24\times 10^{-12}F$

there, energy in the capacitor in parallel combination is given as ;

$E_{P}=\frac{1}{2}C_{P}V^{2}$

$E_{P}=\frac{1}{2}\times 24\times 10^{-12}\times 50\times 50$

$E_{P}=3\times 10^{-8}J$

Related Chapters

Preparation Products

Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Exams
Articles
Questions