CBSE Class 10 Science 2020 Delhi Set 1

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Question : 24
Total: 30
(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 40V.
Solution:  
(a) According to the Joule's law of heating, heat produced in a resistor is directly proportional to the :
(i) square of current (I) for a given resistance.
(ii) resistance (R) for a given current.
(iii) the time ( t ) for which the current. flows through the resistor.
Mathematical form of Joule's law of heating is :
H=I2Rt
(b) Given, charge (q)=96000C
Time (t)=2hrs=120min=7200 s
Potential difference (V)=40 volt
We know that,
Heat (H)=V It, where I is current...(i)
Also,
I=qt, where q is charge and t is time in seconds.....(ii)
By putting I=qt in eqn. (1) we get,
H=(V×qt)×t=Vqtt=Vq
H=40×96000 joule
H=3840000 joule
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