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ICSE Class 10 Physics 2016 Paper

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Question : 21 of 73
Marks: +1, -0
Calculate the quantity of heat produced in a 20Ω20 \Omega resistor carrying 2.5A2.5 \mathrm{A} current in 5 minutes.
Solution:  
Given :I=2.5Amp,R=20Ω,t=5×60=: I = 2.5 \mathrm{Amp}, R = 20 \Omega, t = 5 \times 60 = 300 second.
   Head Produced   =(   Potential Difference   )Volts\; \text{ Head Produced } \; = \underset{\mathrm{Volts}}{\left( \; \text{ Potential Difference } \; \right) } ×(   Current   )Ampere×(   time   )Second\times \left( \; \text{ Current } \; \right)_{\mathrm{ Ampere }} \times \left( \; \text{ time } \; \right)_{\mathrm{ Second }}
=VIt= V I t
=  I2Rt    (   Ohm’s law, V=IR   )= \; I^{2} R t \; \; \left( \; \text{ Ohm's law, V=IR } \; \right)
=    2510×  2510×20×300= \; \; \frac{25}{10} \times \; \frac{25}{10} \times 20 \times 300
=  37500J= \; 37500 \mathrm{J}
=  37.5kJ.= \; 37.5 \mathrm{kJ} .
Hence, 37.5kJ37.5 \mathrm{kJ} heat is produced.
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