Current and inductor

  • Thread starter connor02
  • Start date
  • #1
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Homework Statement



Your circuit has a resistance of 175Ω and an ideal battery with an emf of 6.3V. What value of inductor should you add to the circuit to ensure the current increase in the first 58μs is less than 4.9mA?

Homework Equations



I think

E=L(di/dt)


The Attempt at a Solution



E=L(di/dt)

6.3=L(4.9*10^-3/58*10^-6)

L = 0.07457H

But the answer is 0.069H

Thanks.
 

Answers and Replies

  • #2
I believe you want to use the equation for current growth through a inductor. Also di/dt doesnt equal (4.9*10^-3/58*10^-6)
 
Last edited:
  • #3
27
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WOW you're right! Thanks!!

Then the formula would be I=Io(1-e^(-Rt/L))

Io=6.3/175=0.036A

Solving for L, L=0.069H.

But why is my method wrong? and why is di/dt not (4.9*10^-3/58*10^-6)? di is 4.9*10^-3 and dt is 58*10^-6 right?

Thank you!
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
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When current is flowing, KVL says that the 6.3V is shared (unevenly) between the resistance and the inductance. Your "approach" assumes circuit resistance = 0 and all 6.3V is applied across the inductor.
 
  • #5
27
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Ah, I see, thank you NascentOxygen!!
 

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