Question

a)write the mathematical expression for joule's law of heating
b)compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 v.

Answer 1

Answer:

Hey mate! Hope this helps!

Explanation:

a) A quantitative form of Joule's law is that the heat evolved per second, or the electric power loss, P, equals the current I squared times the resistance R, or P = I2R. The power P has units of watts, or joules per second, when the current is expressed in amperes and the resistance in ohms.

b)Given- Charge(q)= 96000 C

Time(t)= 2 hours=120minutes= 7200 seconds

Potental difference (v)= 40v

we know that,

Heat (H)= vIt, where I is current--eq(1)

Also,

I=q/t, where q is charge and t is time in seconds.-----eq(2)

Therefore by putting I=q/t in eq(1) we get,

H= v*q/t*t

(Here * means multiplication)

Or, H= vqt/t

Or, H=vq

Or, H=40*96000 joule

Or H= 3840000 joule

Thank you!

Answer 2

Answer:

a)H=I²Rt

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b)Given :

Amount of charge transferred =96000C

Time taken =2hrs=2×60×60sec

=7200sec

Potential difference =40V

Heat generated =V×i×t

and we know that ; i=

t

Q

So, H=VQ

=40×96000

=3.84×10⁶J

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