Unformatted text preview: 57 (a) The power factor is cos , where is the phase angle when the current is written i = I sin(d t  ). Thus = 42.0 and cos = cos(42.0 ) = 0.743. (b) Since < 0, d t  > d t and the current leads the emf. (c) The phase angle is given by tan = (XL  XC )/R, where XL is the inductive reactance, XC is the capacitive reactance, and R is the resistance. Now tan = tan(42.0 ) = 0.900, a negative number. This means XL  XC is negative, or XC > XL . The circuit in the box is predominantly capacitive. (d) If the circuit is in resonance, XL is the same as XC , tan is zero, and would be zero. Since is not zero, we conclude the circuit is not in resonance. (e), (f), and (g) Since tan is negative and finite, neither the capacitive reactance nor the resistance is zero. This means the box must contain a capacitor and a resistor. The inductive reactance may be zero, so there need not be an inductor. If there is an inductor, its reactance must be less than that of the capacitor at the operating frequency. (h) The average power is 1 1 Pavg = Em I cos = (75.0 V)(1.20 A)(0.743) = 33.4 W . 2 2 (i) The answers above depend on the frequency only through the phase angle , which is given. If values are given for R, L, and C, then the value of the frequency would also be needed to compute the power factor. ...
View
Full
Document
This homework help was uploaded on 04/15/2008 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.
 Spring '08
 Reich
 Current, Power

Click to edit the document details