Please explain the equations that need to be used... and if you manipulate the e
ID: 3281237 • Letter: P
Question
Please explain the equations that need to be used... and if you manipulate the equation... please write the general equation so I know the basic
equation where you got the final equation from.... thank you so much for your help.
(20%) Problem 2: A V= 114-V source is connected in series with an R = 1.1-k resistor and an L-26-H inductor and the current is allowed to reach maximum. At time t = 0 a switch is thrown that disconnects the voltage source, but leaves the resistor and the inductor connected in their own circuit. After the current decreases to 47 % of its maximum value, the battery is reconnected into the circuit by reversing the switch 50% Part (a) At what value of the time , in milliseconds, does the current reach 79 % of its maximum? Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 20 (0% per attempt) detailed view tan() | | ( cosO cotan0 asin) acos0 sinO HOME atan)acotan) si sinhO 0 coshO tanhO cotanh0 Degrees O Radians END BACKSPACE DE -LEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback. - 50% Part(b) How much energy, in millijoules is supplied in total by the battery, both in initially bringing the current to maximum and in bringing the current back to the 79 % level from 47 %? Ignore energy dissipated in the resistor during those processesExplanation / Answer
Given,
V = 114 V ; R = 1.1 K rom ; L = 26 H
Imax = V/R = 114/1100 = 0.104 A
we know that
I = Im (1 - e^-tR/L)
0.79 Im = Im (1 - e^-tR/L)
e^-tR/L = 0.21
taking natural log both sides
-tR/L = -1.561
t = 1.561 x L/R = 1.561 x 26/1100 = 0.0369s
HEnce, t = 0.0369
b)We know that
U = 1/2 L I^2
47% of Imax = 0.049 A
U = 0.5 x 26 x 0.049^2 = 31 mJ
79% of Imax = 0.082
U = 0.5 x 26 x 0.082^2 = 87 mJ
Umax = 1/2 x 26 x 0.104^2 = 141 mJ
from max to 79%
U1 = 141 - 87 = 54 mJ
U2 = 141 - 31 = 110 mJ