Voltage and current as functions of time for a series RL circuit

In summary, the conversation discusses different approaches to finding the voltage across an inductor. The first method discussed involves using Kirchhoff's laws and solving a differential equation to find the current through the circuit. The second method involves applying KVL to find the potential drop across the inductor, while the third method uses the formula Vc=L(dI/dt) to find the voltage across the inductor. The conversation also mentions the properties of inductors, specifically their behavior when the potential across them changes suddenly and how they behave in a DC circuit.
  • #1
Davidllerenav
424
14
Homework Statement
Deduce the voltage and current as functions of time for the series RL circuit of Figure 1, from the moment the switch SW passes from position 1 to position 2. Explain how you find the value of the inductance Lp using the voltage curve at
the loading or unloading cycle of the coil.
Relevant Equations
Kirchhoff's laws.
##V_L=L\frac{dI}{dt}##
1574719399349.png

I already found ##I(t)## using Kirchhoff's laws, I got the equation ##V-RI-L\frac{dI}{dt}=0\Rightarrow L\frac{dI}{dt}=V-RI## then I solved the differential equation getting ##I(t)=\frac{V}{R}\left[1-e^{-\frac{R}{L}t}\right]##. My problem is founding the voltage as a function of time ##V(t)##, I don't know what to do.
 
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  • #2
Moderator's note: Thread title changed to better reflect the thread content.
 
  • #3
gneill said:
Moderator's note: Thread title changed to better reflect the thread content.
Sorry, I forgot to finish the title.
 
  • #4
If you have I(t), what can you say about the voltage across the resistor? What does KVL have to say about the sum of the potential drops around the circuit?

Another approach might be to note the initial and final conditions for the voltage across the inductor (i.e. at times ##t = 0^+## and ##t → ∞##) and realize that the time constant must be the same as for the current. You should be able to match a standard exponential form to the given information.
 
  • #5
Davidllerenav said:
Sorry, I forgot to finish the title.
No worries. You can always edit your post to fix up things like that, or if the "allowed to edit" time window has closed, ask one of the mentors to help. We're always glad to help!
 
  • #6
gneill said:
If you have I(t), what can you say about the voltage across the resistor? What does KVL have to say about the sum of the potential drops around the circuit?

Another approach might be to note the initial and final conditions for the voltage across the inductor (i.e. at times ##t = 0^+## and ##t → ∞##) and realize that the time constant must be the same as for the current. You should be able to match a standard exponential form to the given information.
KVL says that the sum of potential drops around the equal to zero, becuse it is conserved.

For the second approach, I just need to calculate the limits as ##t\rightarrow 0## and ##t\rightarrow\infty##?
 
  • #7
Davidllerenav said:
KVL says that the sum of potential drops around the equal to zero, becuse it is conserved.
Correct. The drop across the battery is fixed, so that's a known. If you already have ##I(t)## then the drop across the resistance over time should be obvious. That leaves just the drop across the inductance. Apply KVL.

For the second approach, I just need to calculate the limits as ##t\rightarrow 0## and ##t\rightarrow\infty##?
No calculations are really necessary. Given the properties of inductors you should be able to figure out the starting and ending potential drops by inspection.
 
  • #8
gneill said:
Correct. The drop across the battery is fixed, so that's a known. If you already have ##I(t)## then the drop across the resistance over time should be obvious. That leaves just the drop across the inductance. Apply KVL.

So, I just need to find the potential drop across the inductor? I'm not entirely sure I understand why. Applying KVL, I think that the drop across the inductor is ##V_c=L\frac{dI}{dt}##, right?

gneill said:
No calculations are really necessary. Given the properties of inductors you should be able to figure out the starting and ending potential drops by inspection.

I can't see it, what are the properties of inductors?
 
  • #9
Davidllerenav said:
So, I just need to find the potential drop across the inductor? I'm not entirely sure I understand why. Applying KVL, I think that the drop across the inductor is ##V_c=L\frac{dI}{dt}##, right?
The problem statement is frustratingly vague about what voltage you are meant to find. Since they ask about using the voltage curve for the inductor to determine the inductance value, I'm assuming that the inductor voltage is what you need to determine.

You have an expression for the current, so yes, you could use ##V_c=L\frac{dI}{dt}## to find the voltage across the inductor. This is in fact a third method to solve the problem. Well done.

I can't see it, what are the properties of inductors?
What have you learned about how inductors behave when the potential across them changes suddenly (specifically, how the current through them behaves), and how they behave "after a long time" in a DC circuit?
 
  • #10
gneill said:
The problem statement is frustratingly vague about what voltage you are meant to find. Since they ask about using the voltage curve for the inductor to determine the inductance value, I'm assuming that the inductor voltage is what you need to determine.
It makes sense.

gneill said:
You have an expression for the current, so yes, you could use ##V_c=L\frac{dI}{dt}## to vind the voltage across the inductor. This is in fact a third method to solve the problem. Well done.
Thanks. If this is a third method, how would I solve it by the frist method you said then?
gneill said:
What have you learned about how inductors behave when the potential across them changes suddenly (specifically, how the current through them behaves), and how they behave "after a long time" in a DC circuit?
I don't think we have seen much about them. We saw inductance and energy on an inductor.
 
  • #11
Davidllerenav said:
Thanks. If this is a third method, how would I solve it by the frist method you said then?
If you know the potential drops across two of the three components in the loop, KVL allows you to find the third.
I don't think we have seen much about them. We saw inductance and energy on an inductor.
Then you need to know that inductors oppose immediate changes in current. In the instant after a potential difference change is applied to an inductor, it will refuse to change the current through itself. Thus it will produce any required potential difference across itself to enforce this situation in the circuit. Essentially an inductor looks like a fixed current supply in that instant (and yes, zero current is a valid value for that supply, if that's the initial condition).

After "a long time", in the so-called steady-state eventuality, the inductor will behave as a short circuit with no potential difference across it and a constant current through it as determined by the other components in the circuit (in this case the battery and resistance).

Capacitors have an analogous set of properties for initial and eventual behaviors. Capacitors oppose immediate changes in potential difference across themselves and eventually look like open circuits after a sufficiently long time. So they "look like" fixed voltage sources for the instant ##t = 0^+##, and open circuits for ##t → ∞##.
 
  • #12
gneill said:
If you know the potential drops across two of the three components in the loop, KVL allows you to find the third.
So I would have to solve the KVL equation for V?

gneill said:
Then you need to know that inductors oppose immediate changes in current. In the instant after a potential difference change is applied to an inductor, it will refuse to change the current through itself. Thus it will produce any required potential difference across itself to enforce this situation in the circuit. Essentially an inductor looks like a fixed current supply in that instant (and yes, zero current is a valid value for that supply, if that's the initial condition).

After "a long time", in the so-called steady-state eventuality, the inductor will behave as a short circuit with no potential difference across it and a constant current through it as determined by the other components in the circuit (in this case the battery and resistance).

Capacitors have an analogous set of properties for initial and eventual behaviors. Capacitors oppose immediate changes in potential difference across themselves and eventually look like open circuits after a sufficiently long time. So they "look like" fixed voltage sources for the instant t=0+t=0+t = 0^+, and open circuits for t→∞t→∞t → ∞.
Since they behave like a short circuit, I can find the voltage in it, right?
 
  • #13
Davidllerenav said:
So I would have to solve the KVL equation for V?
Yes.
Since they behave like a short circuit, I can find the voltage in it, right?
Yes. You can deduce the voltage as ##t → ∞## from that knowledge. The potential across the inductor will tend to zero as time increases.

If you know the initial voltage and the final voltage and you know that it is an exponential function that is the solution to the problem, you should be able to write the solution essentially by inspection rather than explicitly solving the differential equation. It's a great, practical time saver to know this.
 
  • #14
gneill said:
Yes. You can deduce the voltage as ##t → ∞## from that knowledge. The potential across the inductor will tend to zero as time increases.

If you know the initial voltage and the final voltage and you know that it is an exponential function that is the solution to the problem, you should be able to write the solution essentially by inspection rather than explicitly solving the differential equation. It's a great, practical time saver to know this.
Ok, I will try to do it that way, thanks. About the last part, how would I found the inductance ##L_p## using the voltage curve?
 
  • #15
@gneill I have oone question, when I take the derivative of the current with respecto to time, do I need to take teh derivative of the voltage? Or is it constant? I mean somethiing like this: ##V_L=L\frac{dI}{dt}=L\left(\frac{1}{R}\frac{dV}{dt}\frac{R}{L}e^{-\frac{R}{L}t}\right)##.
 
  • #16
Davidllerenav said:
Ok, I will try to do it that way, thanks. About the last part, how would I found the inductance ##L_p## using the voltage curve?
Essentially you want to find the time constant L/R for the circuit, so that you can determine L.

Given the voltage versus time curve there are several possible approaches. The engineering "rule of thumb" approach is to consider that after five time constants, the circuit has effectively reached steady state. So, look at the curve and decide where the voltage has effectively reached 99% of its final value, and declare that the associated time is five time constants. Use that time value to find the time constant and hence L.

Another approach is to plot ##ln(V)## versus ##t## and look at the slope of the resulting plot (it should be a straight line since the voltage curve follows an exponential function). Think about what the slope of that line tells you.
 
  • #17
Davidllerenav said:
@gneill I have oone question, when I take the derivative of the current with respecto to time, do I need to take teh derivative of the voltage? Or is it constant? I mean somethiing like this: ##V_L=L\frac{dI}{dt}=L\left(\frac{1}{R}\frac{dV}{dt}\frac{R}{L}e^{-\frac{R}{L}t}\right)##.
No, you just need to take the derivative of the current.

##V_L(t) = L \frac{dI(t)}{dt}##

and that's it.

You've already determined ##I(t)##, so you're good to go.
 
  • #18
gneill said:
No, you just need to take the derivative of the current.

##V_L(t) = L \frac{dI(t)}{dt}##

and that's it.

You've already determined ##I(t)##, so you're good to go.
Thanks, I got this answer: ##V_L=Ve^{\frac{-R}{L}t}##. Is it correct?
 
  • #19
Yes, that looks good.
 

FAQ: Voltage and current as functions of time for a series RL circuit

1. What is a series RL circuit?

A series RL circuit is a circuit that contains a resistor (R) and an inductor (L) connected in series. The inductor resists changes in current flow, while the resistor resists the flow of current. This type of circuit is commonly used in electronic devices and power systems.

2. How does voltage change over time in a series RL circuit?

In a series RL circuit, the voltage across the resistor and inductor will vary over time. At first, the voltage will rise as the current starts to flow, but it will eventually level off as the inductor resists changes in current. The voltage across the inductor will also vary depending on the rate of change of current.

3. What is the relationship between current and time in a series RL circuit?

In a series RL circuit, the current will initially be high as the voltage is applied, but it will decrease over time as the inductor resists changes in current. The current will eventually level off as the inductor reaches maximum current flow. The rate of change of current is also related to the voltage across the inductor.

4. How do the values of R and L affect the voltage and current in a series RL circuit?

The value of R (resistance) in a series RL circuit will affect the amount of current that can flow through the circuit. A higher resistance will result in lower current, while a lower resistance will result in higher current. The value of L (inductance) will affect the rate of change of current and the voltage across the inductor.

5. What is the time constant in a series RL circuit?

The time constant in a series RL circuit is a measure of how quickly the circuit will reach steady state. It is calculated by multiplying the resistance (R) and inductance (L) values together. A higher time constant means it will take longer for the circuit to reach steady state, while a lower time constant means it will reach steady state more quickly.

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