Please Explain how to calculate voltage (transients)

[winten]

Lets say we have this circuit:
**broken link removed**
Lets say that it it was still, and at t=1 sec the switch was turned down. Then from 1-5 switch up, and 5-6 switch down again. I calculated currents in this current time:

So, at t=1, I = 5
t=5, I=7.16
t=6, I=6,54

So now how can I calculate inductor voltage for these time moments.

Can i simply use ohms law? How does the graph suppose to look like, does voltage always going down?

And how it is known that voltage of the inductor starts of the 5V?

 

[RCinFLA]

Inductor stores energy in its magnetic field. It takes time for inductor to build up the field. In the inital state of circuit you are changing the amount of stored magnetic energy from an initial steady state of 15v/3 ohms = 5 amps. When switch is closed the current through inductor will ramp up to 15v/2 ohms = 7.5 amps via exponential ramp.

Thoughs are the end points. Other thing to remember is an inductor cannot have its current change instantaneously and a capacitor cannot have its voltage change instantaneously.

Look here for explaination.

 

[MrAl]

Lets say we have this circuit:
**broken link removed**
Lets say that it it was still, and at t=1 sec the switch was turned down. Then from 1-5 switch up, and 5-6 switch down again. I calculated currents in this current time:

So, at t=1, I = 5
t=5, I=7.16
t=6, I=6,54

So now how can I calculate inductor voltage for these time moments.

Can i simply use ohms law? How does the graph suppose to look like, does voltage always going down?

And how it is known that voltage of the inductor starts of the 5V?

Hello there,


The short answer is "Yes", you can use Ohms Law. You have to apply it correctly however, and only to the resistors, but knowing the relationship for the resistor is:
E(t)=I(t)*R
for any time t we can do this for both resistors. Thus, knowing the current at any time t and knowing that the current is the same through every element in a series circuit, we immediately know the voltage drop across any resistor in the circuit for that same time t. This isnt true for the inductor of course, so you can only use this with resistances.

The current itself however is exponential, and to calculate that we have to use i=i0*(1-e^(-t*R/L)) as i am sure you know. Now this doesnt tell us anything about the voltage across the inductor at time t yet, but knowing that the sum of voltage drops around a closed path sums to zero, we know that Vs=v(R1)+v(R2)+v(L), and since we also know Ohms Law we can calculate v(R2) and v(R1) at any time t knowing also I at that time t, so we indirectly know what the voltage across the inductor v(L) is at that time t also.

The calculation could go like this:

1. Calculate i0 by shorting the inductor and calculating the current, this comes out to 5 amps. This is really the current at t infinity though but we call it i0 anyway, keeping in mind that the current at t=0 is zero.
2. Calculate i(t) at t=1, which is:
i(1)=5*(1-e^-3)=4.751065 approx
3. Calculate the voltage drop across the two resistors with total resistance 3 ohms:
v(R)=i(1)*3=14.253195
4. Now that we know the voltage drop across the two resistors, we can calculate the voltage across the inductor by simply subtracting v(R) from the total voltage of 15v:
v(L)=15-14.253195=0.746805 volts approx

The above is an example of how you might calculate these things during the first state of the circuit when the switch is open. I am assuming you know how to solve even after the switch is closed later since the initial conditions for the start of the second state are simply the ending values at the end of the first state.

 

[winten]

Thanks for answers, my teacher said we can use this formula: V=V_0*e^(-t/T), is it possible to use it if we know I at any time, and to get V_0 from ohms law?

 

[MrAl]

Hello again,

Well if your teacher has a specific way he wants you to do this problem then you will really have to ask him for more details.
You can calculate V_0 using Ohms Law in a way knowing that I=0 at t=0, so that means the voltage across the two resistors is 0 at t=0 so:
V_0=15-vR=15-0=15 volts.
You can then use that formula to calculate the voltage some time later.

 

[Ratchit]

winten,

Lets say that it it was still, and at t=1 sec the switch was turned down. Then from 1-5 switch up, and 5-6 switch down again. I calculated currents in this current time:

That sure was a confusing description of the switch settings. First you imply that the switch was open a long time. Then you say that at t=1, the switch was closed. Then you contradict yourself by saying it is open between t=1-5. I am guessing that the switch was initially open and closed from t=1-5, open from t=5-6, and down from t=6 onward.

So, at t=1, I = 5
t=5, I=7.16
t=6, I=6,54

How did you get those values?

So now how can I calculate inductor voltage for these time moments.

See below

Can i simply use ohms law?

No, not simply. You have to use the resistance formula in conjunction with Kirchoff's law and the exponential current change caused by the coil.

How does the graph suppose to look like, does voltage always going down?

What graph? What voltage?

And how it is known that voltage of the inductor starts of the 5V?

Good question. Where do you get 5 volts?

OK, let's start. The formula for current in a coil is derived by differential calculus, but you can use the given resultant formula or consult the universal time-constant chart.

First the currents:
The formula is I = If+(Io-If)*exp(-t/(L/R)) , where If = final current value and Io = initial current value.

T=1 : Switch closes for 4 secs. Io=5.0A,If=7.5A,R=2 . Time constant = 0.5sec, so current will be up to 7.5A well before 4 secs.

T=5 : Switch opens for 1 sec. Io=7.5A, If=5.0A,R=3 . Time constant = 0.3333... Plugging in the values into the current formula above we get exponential decline to I=5.1245A

T=6 : Switch closes indefinitely. Io = 5.1245A,If=7.5A,R=2 . Time constant = 0.5 Current rises exponentially to 7.5A.

Now the voltages:

The formula for voltage across a coil is V = -L(dI/dt) .
T=1 Differentiating the current formula above and multiplying by -L, we get V = -5.0*exp(-2*t) . In 4 secs, the voltage across the coil will be very close to zero. The five volts comes from 5 amps dropping across 2 ohms for -10 volts. The coil produces another -5 volts which matches the 15 volts voltage source. You should be able to figure out the rest of the voltages. Don't forget, when you differentiate the current formula above, the different constants will change the value of the derivative. You will probably find it easier to get the coil voltage by subtracting the voltage drop of the resistors from the voltage source.

Ratch