In the problems below, be methodical, be clear, be neat, and show all your work.

Question

In the problems below, be methodical, be clear, be neat, and show all your work. These are very difficult problems to grade if not done neatly. If I can't understand it, I will have to assume you didn't either. A coil of self-inductance 5.0mH and a resistance of 15.0ohm is placed across the terminals of 12V battery of negligible internal resistance. What is the final current? How much time does it take for the current to reach approximately 63% of its final value? What is the current after 100mus?

Explanation / Answer

the ckt equation for this prblm is
v= Ldi/dt + iR
rearranging we get
di/(V-Ri) = dt/ L
on integration ( at time t=0 , i=0)
we get
i = V/R(1-(e^(-tR/L)))

after sufficiently large enough time the inductor acts as short ckt therfore final current would be V/R= 12/15 =0.8 amps

from above expression time taken for 63% would be
.63V/R= V/R(1-(e^(-tR/L)))
=> t = L/R = 0.33ms

at t= 100s i= .8(1-(e^(-tR/L))) = .207 amps

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