Question

1. A 60-W lightbulb is used in an American desk lamp. (In the United States, standard rms voltage is 120 V.)

(a) What is the average power dissipated by the filament?
.................. W

(b) What is the rms current in the filament?
.................... A

(c) What is the maximum current in the filament?
..................... A

(d) What is the maximum power dissipated by the filament?
....................... W

2.The rms current in a 30.0-Ω resistor connected to an AC source is 2.40 A.

(a) What is the rms voltage across this resistor?
................. V

(b) What is the peak voltage of the AC source?
................ V

(c) What is the average power dissipated by the resistor?
..................... W

(d) What is the maximum current across this resistor?
...................... A

Answer 1

A 60-W lightbulb is used in an American desk lamp. (In the United States, standard rms voltage is 120 V.)

a) 60 W

b) R= V^2/P = 120^2/60 =240 ohm

i_rms = V_rms /R = 120/240 = 0.5 A

c) i_max = i_rms* sqrt(2) = 0.5*sqrt(2) =0.707106781 A

d) P_max = i_max^2*R = =0.707106781^2*240 = 119.96 = 120 W

2.The rms current in a 30.0-Ω resistor connected to an AC source is 2.40 A.

A) V_rms = i_rms *R = 2.4*30 = 72 V

B) V_max = V_rms*sqrt(2) = 72*sqrt(2) =101.823376 V

C) P_avg = V_rms*i_rms = 72*2.4 = 172.8 W

D) i_max = i_rms *sqrt(2) = 2.4*sqrt(2) = 3.39411255 A