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CBSE Class 10 Science Question 17 of 20

Solved 2022 Term 2 Question Paper CBSE Class 10 Science — Question 7

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Question

Question 12(b)

(i) State Joule's law of heating. Express it mathematically when an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t.

(ii) A 5 Ω resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in 10 s.

Answer

(i) Joules law of heating states that the heat dissipated across a resistor is directly proportional to

  • the square of the current flowing through it
  • resistance of the conductor
  • duration of flow of current

i.e., H = I2Rt

(ii) Given,

Resistance (R) = 5 Ω

Voltage (V) = 6 V

Time (t) = 10 s

From ohm's law :

V = IR

6 = I x 5

I = 65\dfrac{6}{5} = 1.2 A

Energy dissipated (H) = I2Rt

Substituting we get,

H = 1.2 x 1.2 x 5 x 10 = 72 J

Hence, energy dissipated = 72 J.