Physics, asked by sankalpsalunke12345, 3 days ago

An electric potential of 100 V is impressed on a certain resistor such that a

current of 12 A is drawn. Calculate the energy dissipated in 3 min.

Answers

Answered by soniatehlan2316

0

Answer:

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Answered by 220513

2

Answer:

0.06KWH

Explanation:

P=v*I

P=100*12

1200watt

1.2KW

time 3min.=3/60hrs

Now, E=p*t

E=1.2*3/60

=0.06Kwh

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