Find Inductor Value to Limit Current Increase in Circuit

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To limit the current increase in a circuit with a resistance of 175Ω and a 6.3V battery to less than 4.9mA in the first 58μs, the correct inductor value is 0.069H. The initial attempt used the formula E = L(di/dt) incorrectly, leading to an erroneous calculation of 0.07457H. The proper approach involves using the equation I = Io(1 - e^(-Rt/L)), where Io is the initial current calculated as 0.036A. The misunderstanding stemmed from assuming that all voltage is applied across the inductor, neglecting the resistance. This highlights the importance of considering circuit components accurately in calculations.
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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.
 
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I believe you want to use the equation for current growth through a inductor. Also di/dt doesn't equal (4.9*10^-3/58*10^-6)
 
Last edited:
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!
 
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.
 
Ah, I see, thank you NascentOxygen!
 
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