Question

12. If a capacitor of capacitance 2 uF is connected
with battery of 10 volt. Then energy used as heat
energy in connecting wires till the capacitor is
completely charged would be [NCERT Pg. 80]
(1) 200 MJ
(2) 100 J
(3) 50 J
(4) 400 J​

Answer 1

Answer:

We know that half the energy supplied by the battery is lost as heat while charging the capacitor it means if a battery supplies the energy EB, then the energy store in the capacitor is 21EB.

Now, from the question

equivalent capacitance =C1+C2C1C2=10+2010×20=200/30=320

Energy stored in this capacitors =21CV2=21×3020×(6)2=10×12=120

Total charge =CV=320×6=40μC

Thus potential drop across the capacitor (C1) is given by V1=1040=4

and potential drop across the capacitor (C2) is given by V2=2040=2

Energy stored in 

Answer 2

Answer:

see the correct answer in attachment

Explanation:

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