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Question

A resistor of resistance 100ohm is connected to an AC source, (12V)sin(250πs1)t. Find the energy dissipated as heat duringt=0to t=1 ms.


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Solution

Given data

  1. The AC voltage source is E=(12V)sin(250πs1)t.
  2. The resistance of the resistor is R=100ohm.
  3. Time duration from t=0to t=1 ms.

Joule's law of heating

  1. If a conductor of resistance carries a current I for time t, then the heat developed in the conductor is proportional to the square of the current.
  2. Joule's law can be defined by the form, H=I2Rt, where, where, H is the heat developed in the conductor.

Calculation of energy dissipation

The given voltage is E=Eosinωt, where, Eo=12volt, ω=250πs-1.

So, energy dissipated in the body is

H=I2Rt=V2R×t(since,V=IR)orH=1R010-3Eo2sin2ωtdt.orH=1100010-3122sin2ωtdt.orH=144100010-3sin2ωtdtorH=1.44010-31-cos2ωt2dtorH=1.442010-3dt-1.44010-3cos2ωt2dtorH=1.442t010-3-1.442sin2ωt2ω010-3orH=0.7210-3-12×250πorH=0.7211000-1500πorH=0.72π-21000=2.61×10-4orH=2.61×10-4joule.

Therefore, the energy dissipated as heat is 2.61×10-4joule.


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