POWER IN A HAIR DRYER
An electric hair dryer is rated at 1500 W (the average power) at 120 V (the rms voltage). Calculate (a) the resistance, (b) the rms current, and (c) the maximum instantaneous power. Assume that the dryer is a pure resistor. (The heating element acts as a resistor.)
IDENTIFY and SET UP:
We are given P_{av} = 1500 W and V_{rms} = 120 V. Our target variables are the resistance R, the rms current I_{rms}, and the maximum value p_{max} of the instantaneous power P.
We solve Eq. (31.29) to find R, Eq. (31.28) to find I_{rms} from V_{rms} and P_{av}, and Eq. (31.30) to find p_{max}.
P_{\mathrm{av}}=\frac{V}{\sqrt{2}} \frac{I}{\sqrt{2}}=V_{\mathrm{rms}} I_{\mathrm{rms}} (31.28)
P_{\mathrm{av}}=I_{\mathrm{rms}}^{2} R=\frac{V_{\mathrm{rms}}^{2}}{R}=V_{\mathrm{rms}} I_{\mathrm{rms}} (31.29)
p=v i=[V \cos (\omega t+\phi)][I \cos \omega t] (31.30)
EXECUTE:
(a) From Eq. (31.29), the resistance is
R=\frac{V_{\mathrm{rms}}^{2}}{P_{\mathrm{av}}}=\frac{(120 \mathrm{~V})^{2}}{1500 \mathrm{~W}}=9.6 \Omega(b) From Eq. (31.28),
I_{\mathrm{rms}}=\frac{P_{\mathrm{av}}}{V_{\mathrm{rms}}}=\frac{1500 \mathrm{~W}}{120 \mathrm{~V}}=12.5 \mathrm{~A}(c) For a pure resistor, the voltage and current are in phase and the phase angle \phi is zero. Hence from Eq. (31.30), the instantaneous power is p=V I \cos ^{2} \omega t and the maximum instantaneous power is p_{max} = VI. From Eq. (31.27) (P_{\mathrm{av}}=\frac{1}{2} V I), this is twice the average power P_{av}, so
p_{\max }=V I=2 P_{\mathrm{av}}=2(1500 \mathrm{~W})=3000 \mathrm{~W}
EVALUATE: We can use Eq. (31.7) (V_{R}=I R) to confirm our result in part (b):
I_{\mathrm{rms}}=V_{\mathrm{rms}} / R=(120 \mathrm{~V}) /(9.6 \Omega)=12.5 \mathrm{~A}.
Note that some unscrupulous manufacturers of stereo amplifiers advertise the peak power output rather than the lower average value.