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

[Davidllerenav]

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.
$V_L=L\frac{dI}{dt}$

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.

 

[gneill]

Moderator's note: Thread title changed to better reflect the thread content.

 

[Davidllerenav]

Moderator's note: Thread title changed to better reflect the thread content.

Sorry, I forgot to finish the title.

 

[gneill]

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.

 

[gneill]

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!

 

[Davidllerenav]

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 $t\rightarrow\infty$?

 

[gneill]

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.

 

[Davidllerenav]

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?

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?

 

[gneill]

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?

 

[Davidllerenav]

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.

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?

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.

 

[gneill]

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 $t → ∞$.

 

[Davidllerenav]

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?

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?

 

[gneill]

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.

 

[Davidllerenav]

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?

 

[Davidllerenav]

@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: $V_L=L\frac{dI}{dt}=L\left(\frac{1}{R}\frac{dV}{dt}\frac{R}{L}e^{-\frac{R}{L}t}\right)$.

 

[gneill]

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.

 

[gneill]

@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.

 

[Davidllerenav]

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?

 

[gneill]

Yes, that looks good.