Class 12 Physics Alternating Current Power associated with resistor

Power associated with resistor

  • The average power over one complete cycle is not equal to 0, the power dissipation is in the form of heat.
  • By using  Joule heating  which is  given as  i2R ;  this shows  Joule heating depends on i2  and  i2 is always positive.
  • This shows there will be average  power consumed by the pure resistive circuits.

Derivation

  • Using the instantaneous values  of voltage and current :-
  • V = Vm sinωt  ,I = Im sinωt 
  • Instantaneous power p = VI
  • Then p = Vm sinωt  Im sinωt 
  • p = Vm Im sin2ωt     equation(1)
  • Average power P = (1/T) 0T∫pdt 
    •  where T = time period
  • =(1/T)  0T∫ Vm Im sin2ωtdt   using equation(1)
  • =( Vm Im/T)  0T∫ (1-cos2ωt)/(2)dt   using (1-cos2θ = 2sin2 θ)
  • P = ( Vm Im/2T)  0T∫(1-cos2ωt) dt
  • =( Vm Im/2T) [t  - (sin2 ωt/2ω)]0T
  • After  simplifying , we will get,
  • P = ( Vm Im/2)
  • Or P =(1/2) Im2 R    equation(2)
  • The above expression shows the pure power consumed in one complete cycle in a pure resistive circuit.
  • In order to represent expression of  power in  AC  same as  in DC,  a new  term was introduced which is  known as  root mean square current.
    • Root mean square current  (rms)  is one of the ways to calculate the average value of alternating current.
    • In order to make the expression of power in  AC  consistent with the expression of power in  DC  the peak  value of  current was replaced by the RMS value.
  • Therefore equation (2) can be written as: - P= I2rmsR.

 

Problem:-  A light bulb is rated at 100W for a 220 V supply. Find (a)  the resistance of the bulb; (b) the peak voltage of the source; and (c)  the rms current through the bulb.

Answer:- 

We are given P = 100 W and V = 220 V. The resistance of the bulb is

R=(V)2/(P)

=(220 )2/(100W) =484

 (b) The peak voltage of the source is

   vm = √2V = 311V

(c) Since, P = I V

I = (P/V) = (100W)/(220V)

=0.450A

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