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 i
^{2}R ; this shows Joule heating depends on i^{2}and i^{2}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 = V
_{m}sinωt ,I = I_{m}sinωt

- Instantaneous power p = VI
- Then p = V
_{m}sinωt I_{m}sinωt **p = V**equation(1)_{m}I_{m }sin^{2}ωt- Average power P = (1/T)
_{0}^{T}∫pdt - where T = time period
- =(1/T)
_{0}^{T}∫ V_{m}I_{m }sin^{2}ωtdt using equation(1) - =( V
_{m}I_{m}/T)_{0}^{T}∫ (1-cos2ωt)/(2)dt using (1-cos2θ = 2sin^{2}θ) - P = ( V
_{m}I_{m}/2T)_{0}^{T}∫(1-cos2ωt) dt - =( V
_{m}I_{m}/2T) [t - (sin2 ωt/2ω)]_{0}^{T} - After simplifying , we will get,
- P = ( V
_{m}I_{m}/2) - Or
**P =(1/2) I**equation(2)_{m}^{2}R - 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
**= I**^{2}_{rms}R.

** 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

v_{m} = √2V = 311V

(c) Since, P = I V

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

=0.450A

.