eAdvanced Physics questic The Expert TAI Hurman11 httpsust3

eAdvanced Physics questic ×/The Expert TAI Hurman-11 , https://ust35nc.theexpertta.com/Common/TakeTutorialAssignment.aspx Class Management Help Practice Problem 11: Inductance, LR & LC Circuits Begin Date: 7/5/2016 11:00:00 AM - Due Date: 8/10/2016 11:00:00 PM End Date: 8/10/2016 11:00:00 PM (1090) Problem 6: Suppose you want the current through a 2.1 H inductor in a series circuit with a 0.45 resistor to decay to 0.100% of its initial value Randomized Variables L=2.1 H R= 0.45 Assignment Status Click here for detailed view 33% Part (a) How much time is required to do this. in seconds? Grade Summa Deductions Potential 096 100% Problem Status Completed Partial cosO cotan0asin acos0 atan acotansinh0 cosh0 tan cotanh0 Subm1ssi0ns Attempts remaining:;3 (4% per attempt) detailed view Partial END Degrees Radians NO Submit Hint I give up! Hints: 49 deduction per hint. Hints remaining: Feedback: 5% deduction per feedback. - 10 33% Part (b) Divide your answer in part (a) by the time constant, L R. and round the result to the nearest integer. That gives you an approximate value for the elapsed time in terms of an integer multiple of the time constant. Now multiply the rounded result by the time constant to obtain the approximate elapsed time in seconds 33% Part (c) By what percent, relative to the more precise value. are the two values for the elapsed tune different? 6:09 PM 8/10/2016 Ask me anything

Solution

For a LR circuit,

The current in the circuit is given by:

I = Io*e^(-tR/L)

where Io = initial current

R = 0.45 ohm

L= 2.1 H

So, for I = Io*0.001

So, 0.001*Io = Io*e^(-t*0.45/2.1)

So, t = 32.2 seconds