Question

a cell of emf 10v and internal resistance 1 ohm is connected a an external resistance of 4ohm then rate of enrgy dissipation in the battery is..

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Answer 1

Explanation:

E=10v

r=1Ω

Rc=4Ω

R=r+Rc

=1+4

=5Ω

I=E/Rc+r

=10/5

=2A

p=I^2R

=2^2×5

=4×5

=20

Pc=P^2×Rc/v^2

=20*4/10. (v=IR=2*5=10v)

=8