In the circuit shown in the figure, the switch S is in position 1 for a long time. Now the switch is thrown to position 2. Find the total heat developed in the circuit till the steady state is reached again

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

0.2 J

0.4 J

0.02 J

0.04 J

answer is A.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Initial charge on capacitor: $Q_{1} = 2 \times 100 = 200 μ C$

Final charge: $Q_{2} = 2 \times 300 = 600 μ C$

$W_{b} = 300 [Q_{2} - Q_{1}] = 12, 000 μ J$

$Δ U = \frac{1}{2} C [300^{2} - 100^{2}] = 80000 μ J$

Heat $= W_{b} - Δ U = 40000 μ J = 0.04 J$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

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

Take Free Test