Sprint LTE e * 42% 7:03 PM triton.physics.ndsu.nodak.edu Course Contents . » Hom

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Sprint LTE e * 42% 7:03 PM triton.physics.ndsu.nodak.edu Course Contents . » Homework 10 » KL Circuit Timer Notes Evaluate FeedbackPrint Info Consider an RL circuit as shown in the Figure with V V 20 V, L - 30 mH, and R- 1500 2. After the switch has been left open for a long time, the switch is closed att 0 (a) What is the current the instant after the switch is closed? Submit AnswerTries 0/5 (b) What is the current a very long time after the switch is closed? Submit AnswerTries 0/5 (c) How long does it take the current in the circuit to reach 40 % of its final (long time) value? Submit AnswerTries 0/5 Post Discussion Send Feedback

Explanation / Answer

part A :

equation for current in RL ckt is

I = Io (1- e^-Rt/L)

at an insant t = 0

I = Io* (1- e^0)

I = 0

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after a very long time , Current is steady

I = V/R = 20/1500

I = 0.013 AMps

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for reaching 40%

use I = Io (1- e^-Rt/L)

0.4(I/Io) = 1-e^-Rt/L

e^-Rt/L = 1-0.4 = 0.6

-Rt/L = ln(0.6)

Rt/L = -0.51

t = 0.51 * 30*10^-3/(1500)

t = 10.2 micro secs

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