The power absorbed by a non-linear device is p = 9(e.16t² – 1) . if3(e0.4t + 1), how much charge goes through this device between0

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The power absorbed by a non-linear device is p = 9(e0.16t² – 1) . if v = 3(e04t + 1), how much charge goes through this device between 0st<9 seconds? A) 8.400 C E) 2.973 kC B) 57.173 C F) 9.984 kC C) 239.986 C G) 33.282 kC D) 867.828 C H) 110.656 kC

The power absorbed by a non-linear device is p = 9(e0.16t² – 1) . if v = 3(e04t + 1), how much charge goes through this device between 0st<9 seconds? A) 8.400 C E) 2.973 kC B) 57.173 C F) 9.984 kC C) 239.986 C G) 33.282 kC D) 867.828 C H) 110.656 kC

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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HVDC

HVDC stands for high voltage direct current in the electric power transmission system. The high voltage direct current system uses direct current (DC) for the transmission of electrical power through transmission lines in contrast with the alternating current that is the AC system. The HVDC is used to transmit bulk power point-to-point through long distances transmission lines. Power is rapidly transferred because the efficiency of transmission lines is greatly increased over long distances due to HVDC lines.

Question

The power absorbed by a non-linear device is p = 9(e.16t² – 1) . if
3(e0.4t + 1), how much charge goes through this device between
0<t<9 seconds?
V =
A) 8.400 C
E) 2.973 kC
B) 57.173 C
F) 9.984 kC
C) 239.986 C
G) 33.282 kC
D) 867.828 C
H) 110.656 kC

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