Question

hey guys can u plz help me .. This is class 10 cbse question (1)compute the heat generated while transferring 96000 coloumb of charge in one hour through a potential difference of 50 V​

Answer 1

Answer:

4.8 x 10^6 J

Explanation:

Charge (Q) = 96000 C

Time (t) = 1hr = 60 x 60 = 3600 s

Potential difference (V) = 50 V

I = Q/t

I =  (96000)/(3600)

I = (80/3) A

 H = V × I × t

H = 50 × (80/3) × 3600

H = 4.8 × 10^6 J

Therefore, the heat generated is 4.8 x 10^6 J.

Answer 2

Answer:

1333.3W

Explanation:

Q=96000 C

t=1 hr=3600seconds

I=tQ=360096000=380A

V=50 V

Amount of heat generated is 

H=VIt

=50×380×3600

=4800000 J

Power input by the source is 

P=VI

=50×380

=1333.3 W