A capacitor of capacitance $10 μ F$ is charged up to a potential difference of $2 V$ and then the cell $s$ removed. Now it is connected to a cell of emf $4 V$ and is charged fully. In both cases the polarities of the two cells are in the same directions. Total heat produced in the complete charging process in $μ J$ is ____

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 20.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Energy stored in capacitor when it is charged upto
$2 V = \frac{1}{2} 10 2^{2} = 20 μ J$
Energy stored in capacitor when it is charged upto
$4 V = \frac{1}{2} 10 4^{2} = 80 μ J$
Increase in charge $= 40 - 20 = 20 μ C$
Energy drawn from cell $= 20 \times 4$
$= 80 μ J = μ (\sup p o s e)$ Heat produced
$= μ_{1} = μ_{1} + μ - μ_{2} = 20 + 80 - 80 = 20 μ J$

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET