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

In summary, the conversation discusses the connection of an inductor to a battery with an emf of 12.0V and negligible internal resistance. The initial current is 4.86mA and after a long time, it is 6.45mA. The attempt at a solution involves finding the resistance and inductance of the circuit, but the solution is not accepted due to neglecting the presence of the battery.
  • #1
xSpartanCx
18
0

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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  • #2
Derp. should've been using 1- exp(-t/(L/R))
 
  • #3
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.
 

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

What is an LR circuit?

An LR circuit is a type of electrical circuit that consists of a resistor (R) and an inductor (L) connected in series. The inductor stores energy in the form of a magnetic field, while the resistor dissipates energy as heat.

What is inductance in an LR circuit?

Inductance is the property of an inductor that resists changes in current flow. In an LR circuit, inductance is measured in units of henries (H) and is represented by the symbol L. It determines how much energy is stored in the magnetic field of the inductor.

How does inductance affect the behavior of an LR circuit?

Inductance plays a crucial role in the behavior of an LR circuit. When the circuit is first energized, the inductor resists changes in current, causing a delay in the flow of current through the circuit. As the current continues to flow, the inductor's magnetic field builds up, storing energy. When the power source is removed, the inductor releases this stored energy, causing a brief surge of current through the circuit.

What is the time constant in an LR circuit?

The time constant in an LR circuit is a measure of how quickly the inductor's magnetic field builds up or decays. It is calculated by multiplying the inductance (L) by the resistance (R) in the circuit. The time constant is expressed in units of seconds (s).

What are some real-world applications of LR circuits?

LR circuits have many practical applications, including in power supplies, electronic filters, and electric motors. They are also used in audio equipment to remove unwanted frequencies, and in radios to tune in specific frequencies. Additionally, LR circuits are commonly found in electronic ballasts for fluorescent lights, which use inductance to limit the amount of current flowing through the bulb.

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