Question

Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 v

Answer 1

Given- Charge(q)= 96000 C

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

Potential 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

Therefore the heat generated while transferring 96000 coloumb of charge in two hours through a potential difference of 40v is 3840000 joules.

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Answer 2

Answer:

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

Step-by-step explanation: