Question

calculate the energy transferred by 5 ampere current flowing through resistors of 2 ohm in 30 minutes

Answer 1

Energy transferred = I²*R*t

5²*2*30*60

= 90kJ

Answer 2

Answer:I=5A

R=2 ohms

t=30min=1/30 he

Now by,

P=I R

P=25×2=50w

Then,

E=P×t

E=50×1/60

=5/6

=0.83kwh

Explanation: