Differential Difficulties in an RL Circuit Problem

In summary: Now, if you add the voltage drops around the loop, they should sum to zero (Kirchhoff's voltage law). You get a DE and express the solution in terms of initial conditions. So, for your problem, you can write$$L\frac{di}{dt} + R\left(i - \frac{v}{R}\right) = 0$$where v is the applied potential difference. You need to solve this equation and then find di/dt at t = 1.2 ms.
  • #1
Geromy
8
0

Homework Statement



An inductor with L = 50 mH is in series with a resistor of R = 180 ohms. At t = 0, a potential difference of 45 V is suddenly applied across the series circuit. At what rate is the current increasing after 1.2 milliseconds?



Homework Equations



V = IR + Emf
Emf = -L(dI/dt)
I = dQ/dt


The Attempt at a Solution



My first instinct was to try and find the maximum current, or when the rate of change of the current is 0, but I'm not sure that doing that would accomplish anything. I've been puzzling over how to relate the concepts with differential equations, but I've gotten pretty stuck there, too. Any points in the right direction would be greatly appreciated! Thank you!
 
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  • #2
First of all, you need to construct the DE which describes this circuit and solve it to obtain I(t). Once you do that, then you can calculate dI/dt for t = 1.2 ms.
 
  • #3
Okay! But I'm kind of lost there, too - what information do I need to construct the differential equation?
 
  • #4
Geromy said:
Okay! But I'm kind of lost there, too - what information do I need to construct the differential equation?

KCL would be a good start.
 
  • #5
I don't know anything about KCL...

After doing a little research, though, it looks like maybe Q(t) = L*V(1-e^(tR/L)): if this is true, then it's just a matter of derivatives. I'm not sure it's right, though (I cobbled it together from Wikipedia and Hyperphysics) - would knowing the KCL help?

(the formula comes from a similar one which I learned for RC circuits, in which Q(t) = C*V(1-e^(t/RC)), so I'm not so sure that it's as simple as switching the time constant)
 
  • #6
Geromy said:
I don't know anything about KCL...

After doing a little research, though, it looks like maybe Q(t) = L*V(1-e^(tR/L)): if this is true, then it's just a matter of derivatives. I'm not sure it's right, though (I cobbled it together from Wikipedia and Hyperphysics) - would knowing the KCL help?

(the formula comes from a similar one which I learned for RC circuits, in which Q(t) = C*V(1-e^(t/RC)), so I'm not so sure that it's as simple as switching the time constant)

I don't remember the solutions, its quite easy to find them. I meant Kirchoff's laws, you should have heard of them.
 
  • #7
Ah! Yes, okay - I think that's already been essentially completed. The circuit is a single loop in series, so V = IR + Emf, like I put above.
 
  • #8
Geromy said:
Ah! Yes, okay - I think that's already been essentially completed. The circuit is a single loop in series, so V = IR + Emf, like I put above.

Yes. :)

So what is the D.E (differential equation) you get?
 
  • #9
V = R(dQ/dt) - L(dI/dt)

The problem is I don't know the equation describing Q - I've found a candidate, but I'm not terribly confident about it.
 
  • #10
Geromy said:
V = R(dQ/dt) - L(dI/dt)

The problem is I don't know the equation describing Q - I've found a candidate, but I'm not terribly confident about it.

Why minus? :confused:
 
  • #11
The back Emf opposes the direction of current flow, so it's negative until the current starts switching directions, is my understanding
 
  • #12
Geromy said:
The back Emf opposes the direction of current flow, so it's negative until the current starts switching directions, is my understanding

Well, I am not sure how to explain it as I am a student myself, not an expert. :)

Yes, the back emf opposes the direction of current flow. As you state, it's negative which would mean,

V-L(dI/dt)=IR

Agreed?
 
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  • #13
Geromy said:
V = R(dQ/dt) - L(dI/dt)

The problem is I don't know the equation describing Q - I've found a candidate, but I'm not terribly confident about it.
What is Q supposed to represent in the circuit? The charge of what? The answer is there is no element in the circuit with a charge Q, so it doesn't make sense to express any quantity in terms of this ill-defined variable.

The current I is apparently the current through the inductor. How is that related to the current through the resistor?
 
  • #14
Geromy said:
The back Emf opposes the direction of current flow, so it's negative until the current starts switching directions, is my understanding
From a problem-solving standpoint, you should follow this convention: Assume a direction of current flow for each element (R, L, and C). Label the end where the current enters with a + and the end where the current leaves with a -. As you traverse a loop, if you go from + to - across an element, you subtract the potential difference. If you go from - to +, you add the potential difference. With this sign convention, you have the relations
\begin{align*}
v &= iR \\
i &= C\frac{dv}{dt} \\
v &= L\frac{di}{dt}
\end{align*} where v is the potential difference across the element and i is the current flowing through the element.

(Batteries follow the opposite sign convention because they are sources. The current flows out through the + side.)
 

1. What is an RL circuit?

An RL circuit is an electrical circuit that consists of a resistor (R) and an inductor (L) connected in series. This type of circuit is commonly used in electronic devices and systems.

2. What causes differential difficulties in an RL circuit problem?

Differential difficulties in an RL circuit problem can be caused by a sudden change in the current or voltage in the circuit. This can lead to the generation of a back EMF (electromotive force) in the inductor, which can affect the overall behavior of the circuit.

3. How does an inductor affect the behavior of an RL circuit?

An inductor is a passive electronic component that stores energy in the form of a magnetic field. When a sudden change in current occurs in the circuit, the inductor resists this change by generating a back EMF. This can cause difficulties in solving RL circuit problems as it affects the overall behavior of the circuit.

4. How can I solve differential difficulties in an RL circuit problem?

To solve differential difficulties in an RL circuit problem, you can use techniques such as Kirchhoff's laws, Ohm's law, and the voltage-current phase relationship for resistors and inductors. Additionally, you can use differential equations to model the behavior of the circuit and find the solution.

5. What are some applications of RL circuits?

RL circuits have various applications in electronic devices and systems, including power supplies, electric motors, generators, and filters. They are also commonly used in communication systems, such as radio and television receivers, to tune the frequency of the incoming signal.

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