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