Question

 

A. According to Equation 20.7, an ac voltage V is given as a function of time t by V = V_o sin 2ft, where V_o is the peak voltage and f is the frequency (in hertz). For a frequency of 56.0 Hz, what is the smallest value of the time at which the voltage equals one-half of the peak-value?

B. The rms current in a copy machine is 7.36 A, and the resistance of the machine is 19.8Ω. What are (a) the average power and (b) the peak power delivered to the machine?

C. A portable electric heater uses 21.8 A of current. The manufacturer recommends that an extension cord attached to the heater receive no more than 2.64 W of power per meter of length. What is the smallest radius of copper (resistivity 1.72 x 10^-8 Ω·m) wire that can be used in the extension cord? (Note: An extension cord contains two wires.)   of about 12A

D. The average power used by a stereo speaker is 65 W. Assuming that the speaker can be treated as a 4.3-Ω resistance, find the peak value of the ac voltage applied to the speaker.

 

Answer 1

A) Te peak voltage is V = Vo sin 2f t

to have V one half of the peak value

we have V= Vo /2

so

solving futher we get t = 4.67 ms

B) The avergae current Ia = 7.36 A and resistance R = 19.8 ohm

The average power is

The peak current is Ip = 1.414 Ia = 1.414 X 7.36 = 10.41 A

The peak power is

C) The current in I = 21.8 A , The power per one meter is P = 2.64 W ,

as Power P = I²R so , resistance R = 5.55 X 10-3 Ohm for 1 m length

as R is

where is the resistivity = 1.72 x 10^-8 Ω·m , L = 1 m R = 5.55 X 10-3

putting the values and solving we get r = 9.93 X 10 -4 m or 0.993 mm

D) The average power is given by

where Pa averge power Pa = 65W , Resistance R = 4.3 omh , and average ac Voltage is Va

solving further we get Va = 3.89 V

The peak voltage is Vp = 1.414 Va = 1.414 X 3.89 = 5.49 V