Class Management |Help Practice Problem 11: Inductance, LR & LC Circuits Begin D
ID: 1787406 • Letter: C
Question
Class Management |Help Practice Problem 11: Inductance, LR & LC Circuits Begin Date: 4/112016 12:00:00 AM - Due Date: 12/8/2017 2:00:00 AM End Date: 12/8/2017 2:00:00 AM (10%) Problem 2: An RL circuit is shown on the right. L = 2.8 H. R = 106 , = 5.5 V ©theexpertta.com 9% Part (a) At moment 1-0, close the switch. Express the current l in the circuit as a function of time in terms of L. R, and Grade Summary Deductions Potential 0% 100% tR Submissions Attempts remaining: 5 (2000 per attenpt) detailed view tR Submit lint I give up! Hints: 4% deduction per hint. Hints remaining: 3 Feedback: 5% deduction per feedback. 9% Part (b) Express the inductive timc constant, , in terms ofL and R 9% Part (c) Calculate the numerical value of t 9% Part (d) Calculate the numerical value of 1 at t-1 s in A 9% Part (e) Express the time th when the current reaches one half of its maximum value, in terms of L and R 9% Part (f) Calculate the numerical value of 10 in s 9% Part (g) What's the direction of the current, clockwise or counterclockwise? 906 Part (h) Express the magnitude of the induced emfon the inductor, ., in terms ofL and 1 9% Part (i) Express ,, as a function of time in terms of e, L, and R 9% Part (j) Calculate the numerical value of,L at 0.01 s in V 9% Part (1) What's the direction of the induced current L counterclockwise or clockwise?Explanation / Answer
2. L = 2.8 H
R = 106 ohms
e = 5.5 V
a. at t = 0 , switch is closed
then
from kirchoff's law
e = Ldi/dt + iR
Ldi/dt = e - iR
Ldi/(e - iR) = dt
integrating from t = 0 to t = t
L*[ln(e - iR) - ln(e)] = -Rt
ln(1 - iR/e) = -Rt/L
1 - iR/e = e^(-Rt/L)
i = (e/R)(1 - e^(-Rt/L))
so option 1.
b. t = L/R
c. t = 2.8/106 = 0.0265s
d. at t = 1 s
i = (5.5/106)(1 - e^(-106*1/2.8)) = 0.051886 A
e. for half current
0.5 = 1 - e^(-106*to/2.8)
to = 0.0183 s