What is the voltage across an inductor in an LR circuit?

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SUMMARY

The voltage across an inductor in an LR circuit is determined by the relationship between the inductor's resistance and the applied voltage from the battery. In this case, a 12.0V battery is connected to an inductor with a resistance calculated to be 1860 Ohms. The current at steady state is 6.45mA, while at 0.940ms, the current is 4.86mA. The inductance is calculated to be 6.18 H, using the equation V(t) = V0(1 - exp(-t/(L/R))) to account for the voltage across the inductor.

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  • Understanding of LR circuits and their behavior over time
  • Familiarity with Ohm's Law (V = IR)
  • Knowledge of exponential decay in electrical circuits
  • Ability to manipulate natural logarithms for solving equations
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  • Learn about the time constant (τ) in LR circuits and its significance
  • Explore the application of Kirchhoff's laws in analyzing circuit voltages
  • Investigate the differences between ideal and real inductors in practical applications
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xSpartanCx
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Homework Statement


An inductor is connected to the terminals of a battery that has an emf of 12.0Vand negligible internal resistance. The current is 4.86mA at 0.940ms after the connection is completed. After a long time the current is 6.45mA.

Homework Equations


V = IR
V(t) = V0 (exp(-t/τ))

The Attempt at a Solution


At infinite time, an inductor appears like a wire. Therefore, 12V = 6.45mA * R
R = 1860 Ohms

V(t) = V0 exp(-t/(L/R))
(4.86mA * 1860 Ohm) = 12V exp(-t/(L/R))
.753 = exp(-t/(L/R))
natural log both sides...
.28 = t / (L/R)
L/R = 0.0033
L = 6.18 H

However, MasteringPhysics does not accept this solution.
 
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Derp. should've been using 1- exp(-t/(L/R))
 
xSpartanCx said:

Homework Statement


An inductor is connected to the terminals of a battery that has an emf of 12.0Vand negligible internal resistance. The current is 4.86mA at 0.940ms after the connection is completed. After a long time the current is 6.45mA.

Homework Equations


V = IR
V(t) = V0 (exp(-t/τ))

The Attempt at a Solution


At infinite time, an inductor appears like a wire. Therefore, 12V = 6.45mA * R
R = 1860 Ohms

V(t) = V0 exp(-t/(L/R))
What is V(t)? The voltage across what?

xSpartanCx said:
(4.86mA * 1860 Ohm) = 12V exp(-t/(L/R))
.753 = exp(-t/(L/R))
natural log both sides...
.28 = t / (L/R)
L/R = 0.0033
L = 6.18 H

However, MasteringPhysics does not accept this solution.

You completely ignored the battery. It is present. You can consider the coil as an ideal inductor connected in series with its resistance and connected to the battery. The voltage across the inductor is not the same as the voltage across the resistor.
 

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