# The potential difference/EMF in RL circuits

• hidemi
In summary, the inductance in a circuit will decay exponentially, reaching zero over time. The potential difference across the resistor will be close to zero.
hidemi
Homework Statement
(1) An inductance L, resistance R, and ideal battery of emf ε are wired in series and the circuit is allowed to come to equilibrium. A switch in the circuit is opened at time t = 0, at which time the current is ε/R. At any later time t the potential difference across the resistor is given by:

A.ε (1−e^−Lt/R)
B.εe^−Lt/R
C.ε (1 +e^−Rt/L)
D.εe^−Rt/L
E. ε (1−e^−Rt/L) The answer is D.

(2) An inductance L, resistance R, and ideal battery of emf ε are wired in series. A switch in the circuit is closed at time t = 0, at which time the current is zero. At any later time t the emf of the inductor is given by:

A.ε (1−e^−Lt/R)
B.εe^−Lt/R
C.ε (1 +e^−Rt/L)
D.εe^−Rt/L
E. ε (1−e^−Rt/L) The answer is D.
Relevant Equations
Potential Difference = ε*e^(-Rt/L)
I would like to share my understanding of both of these questions.

For Question (1), the inductance L is releasing its stored energy to be dissipated by the resistance R. As time passes, the voltage across L is decreasing and thus the potential difference across the resistor will be close to 0.

For Question (2), the inductance L is being charged, as the current is given zero initially and thus voltage is at its maximum (because the inductance has a maximum reactance in the beginning). As time passes, the emf of the inductance drops (less reactance), so the emf of the inductor is close to 0.

Let me know if my thought is correct.

For (1), is there a diagram for the circuit in question?

If it is purely a series circuit and an inline switch is opened, where's the closed circuit to allow current to flow once that switch is opened?

Steve4Physics
gneill said:
For (1), is there a diagram for the circuit in question?

If it is purely a series circuit and an inline switch is opened, where's the closed circuit to allow current to flow once that switch is opened?
There's no diagrams provided in both Question (1) and (2), but I could draw the diagrams from my own understanding. (See attachment)

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hidemi said:
There's no diagrams provided in both Question (1) and (2), but I could draw the diagrams from my own understanding. (See attachment)
The descriptions in the Homework Statement aren’t quite correct. I think this is intended:
https://s3-us-west-2.amazonaws.com/...ites/222/2014/12/20110934/Figure_24_10_01.jpg
(from https://courses.lumenlearning.com/physics/chapter/23-10-rl-circuits/)

For Q1, switch is in position 1, circuit reaches equilibrium. At t=0 switch is changed to position 2.
For Q2, switch is in position 2, circuit reaches equilibrium. At t=0 switch is changed to position 1.

hidemi
Steve4Physics said:
The descriptions in the Homework Statement aren’t quite correct. I think this is intended:
https://s3-us-west-2.amazonaws.com/...ites/222/2014/12/20110934/Figure_24_10_01.jpg
(from https://courses.lumenlearning.com/physics/chapter/23-10-rl-circuits/)

For Q1, switch is in position 1, circuit reaches equilibrium. At t=0 switch is changed to position 2.
For Q2, switch is in position 2, circuit reaches equilibrium. At t=0 switch is changed to position 1.
Thanks for the response!
I think your diagrams make more sense.
If those are the cases, then do my thoughts on post#1 still remain valid?

hidemi said:
Thanks for the response!
I think your diagrams make more sense.
If those are the cases, then do my thoughts on post#1 still remain valid?
Your explanation for Q1 is OK but (being picky) rather than say
“the potential difference across the resistor will be close to 0.“
I would say something like
“the potential difference across the resistor decays exponentially (tending to zero for large times)”

For Q2 you said
“thus voltage is at its maximum (because the inductance has a maximum reactance in the beginning)”.
That's wrong.

You can’t use the term ‘reactance’ here because the concept of reactance is used for alternating current flowing through an inductor (inductive reactance ##X_L = 2\pi f L## wherre ##f## is the frequency).

I would say somethng like:
“thus emf is at its maximum (because the current is changing fastest at the beginning and induced emf is proportional to rate of change of current)”

hidemi
Steve4Physics said:
Your explanation for Q1 is OK but (being picky) rather than say
“the potential difference across the resistor will be close to 0.“
I would say something like
“the potential difference across the resistor decays exponentially (tending to zero for large times)”

For Q2 you said
“thus voltage is at its maximum (because the inductance has a maximum reactance in the beginning)”.
That's wrong.

You can’t use the term ‘reactance’ here because the concept of reactance is used for alternating current flowing through an inductor (inductive reactance ##X_L = 2\pi f L## wherre ##f## is the frequency).

I would say somethng like:
“thus emf is at its maximum (because the current is changing fastest at the beginning and induced emf is proportional to rate of change of current)”

## 1. What is the potential difference/EMF in an RL circuit?

The potential difference, also known as electromotive force (EMF), in an RL circuit is the voltage difference between the positive and negative terminals of the power source.

## 2. How is the potential difference/EMF calculated in an RL circuit?

The potential difference/EMF in an RL circuit can be calculated using Ohm's Law, which states that the voltage (V) is equal to the product of the current (I) and the resistance (R): V = I x R. In an RL circuit, the resistance is the total resistance of the circuit, which includes the resistance of the resistor (R) and the inductor (L).

## 3. What factors affect the potential difference/EMF in an RL circuit?

The potential difference/EMF in an RL circuit is affected by the resistance of the circuit, the inductance of the inductor, and the frequency of the alternating current.

## 4. How does the potential difference/EMF change over time in an RL circuit?

In an RL circuit, the potential difference/EMF initially increases as the current begins to flow, but then decreases over time as the inductor builds up a magnetic field. Once the magnetic field is fully established, the potential difference/EMF reaches a steady state and remains constant.

## 5. What is the significance of the potential difference/EMF in an RL circuit?

The potential difference/EMF in an RL circuit is important because it determines the rate at which the current changes in the circuit. It also plays a crucial role in the behavior of inductors, which are used in many electronic devices such as motors and generators.

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