C A Charged Belt, 50 Cm VxWileyPLUS C Chegg Study l Guided Sx 900w to kw - Googl
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C A Charged Belt, 50 Cm VxWileyPLUS C Chegg Study l Guided Sx 900w to kw - Google Se X Secure https://edugen.wleyplus.com/edugen/st Apps It. We Went And UBITName Logic A WileyPLUS Halliday, Fundamentals of Physics, 100 University Physics (PHYS 107/ PHYS108/ PHYS 207) Home Read, Study & Practice Gradebook ORION Downloadable eTextbook Assignment> Open Assignment MESSAGE MY INSIRUCTOR FULL SCREEN PRINTER VERSION BACK NEXT ASSIGNMENT RESOURCES Chapter 26, Problem 085 Summer 2017 PHY108 CH 26 Chapter20·Problem A 38 F capacitor is connected across a programmed power supply. Dunng the interval from-0 to t = 4.00 s the output voltage of the supply is given by V(t) = 6.00 + 4.00t. 2.00r2 volts. At-0.500 s f find (a) the charge on the capacitor, (b) the current into the capacitor, and (c) the power output from the power supply. 003 006 019 024 040 241 Units Units Units (a) Number Chaoter 20·Problem a Qhaoter 26 Prablem(b) Number a Chater 26, Problem c) Number LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEMVIDEO MINT-LECTURE Review Score Question Attempts: 0 of 4 used SAVE FOR LATER SUBMIT ANSWER All Rights Reserved. A Division of John Wley &Sons; Inc Version 4.24.0.3 O Type here to search 7/30/2017 4Explanation / Answer
(a) We know that the charge on the capacitor is given by
q = CV
Hence at t = 0.5
V(t) = 6 + 4*(0.5) -2(0.5)2 = 7.5 V
hence charge
q = 38*10-6*(7.5) = 285*10-6 C = 285 uC
(b) We know that the current is
I = dq/dt = d(CV)/dt
I = C*(dV/dt) = C*(0+4 -4t) = C(4- 4t)
I = 38*10-6*(4- 4*0.5) = 76*10-6 A
(C) Power of the capacitor is given by
= VI
At t= 0.5 , V = 7.5 V (calculated earlier)
I = 76*10-6
P = 76*10-6*7.5 = 570*10-6 W
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