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.
- 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
- =(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
=(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)