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.

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

.