Question

3. A current of 4 amp flows in a wire of resistance 60 С2. Calculate:
(a) Power dissipated in the wire.
(b) Potential difference across the wire.
(c) Electrical energy consumed in 2 minutes.​

Answer 1

Answer:

a) P=(I×I)×R

given: I=4A,R=60

Power=4×4×60

= 960watt

b)V=IR

=4×60

=240volts

c)E=(V×V/R)×T

=(240×240/960)×60×2

=7200joule

Answer 2

Answer:

(i)960W  (ii)240V  (iii)7200J

Explanation:

current=4 amp

resistance=60 ohms

therefore,by ohm's law,

          we have, V=IR

                           V=60*4=240 V

power=V*I

               p=240*4=960 W

 electrical energy=(V*V/r)*t

             2mins=120sec

             E=(240*240/60)*120

             E=7,200 J.