Verizon ITE 8:53 PM 59% phy-loncapa.cst.cmich.edu Show Metadata Grepare a docume

Question

Verizon ITE 8:53 PM 59% phy-loncapa.cst.cmich.edu Show Metadata Grepare a document nones and annotations PHY131 S17A A 10 The figure below shows the time variation ofthe current through an electrical heater when it is plugged into a 120V50 z outlet. What is the peak voltage? What is the mms value of the current drawn by the heater? Tries 020 What is the resistance of the heater? ft -0.02 s, what is the value of t2? Note: The graph does not start at t-0 seconds. Threaded View ChronologicaLView SortingEiltering options Export? NEW Part 4 Anonymous Reply Tue Mar l4 12:02:35 pm 2017 (EDT) Does anyone know how to find part 4 of this problem? NEW Re: Part 4 Brad Harris harrisba:cmich) Reply Tue Mar 14 07:26:29 pm 2017 (EDT) you use the seconds given to us the add (cycles/frequency) Re: Re: Part 4 Brad Harris (harri5ba:cmich) Reply Tue Mar 14 07.26:53 pm 2017 NEW then. NEW help Anonymous 3 Reply Tue Mar l408:50:52 pm 2017 (EDT) I still don't understand part 4 NEW Anonymous Reply (Wed Mar 15 10:44:37 am 2017 (EDT) Im having trouble figuring out the current, dont we need to know the power or resistance to do that?

Explanation / Answer

Imax = 10 A
Vrms = 120 V
Vpeak = Vrms * sqrt(2)
VPeak = 120 * sqrt(2)
Vpeak = 169.7 V

Irms = Imax/sqrt(2)
Irms = 10 / sqrt(2)
Irms = 7.07 A

Vrms = Irms * R

120 = 7.07 * R
R = 17

Given,
f = 50 Hz
t1 = 0.02 s
t = 1/f = 1/50 = 0.02 s

t2 = t1 + 1.5 * t
t2 = 0.02 + 1.5 * 0.02
t2 = 0.05 s

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.