# Help Understanding zero state RL circuit *first derivative*

#### dacrazyazn

Joined Mar 30, 2009
22
I've been at this problem for awhile. looked at my notes in class, and the book and it is still kinda confusing to wrap my head around.

Basically, i need help understanding first derivative RL circuit with zero input, So the is power in a capacitor or an inductor and this has to do with something... As time reaches infinite, i know that it'll go to zero, but what is the point?

more specifically, i have a problem of which the power source (Vs), Inductor (L) and the resistor (R) are all in series, and the voltage is taken from the two sides of the resistor (Vo) and the question asks..

"the circuit is in the ZERO STATE when the input Vs(t)=Va*u(t) V is applied. Find Vo for t>=0. Identify the forced and natural components in the output."

I feel that if i know the basic concept of the problem, and WHAT the problem is asking for, i can solve it on my own. I especially dont understand it when it saids... ZERO STATE. what does that mean?

#### jasperthecat

Joined Mar 26, 2009
20
dacrazyazn

The zero state is setting the initial conditions for the circuit (and the subsequent equations) - ie nothing is happening - no current flowing - no rising or falling magnetic field in the inductor.

The question is asking you derive and solve the differential equation for the series R & L circuit when Vs(t)=Va*u(t) is applied - specifically looking at the voltage across the resistor. Do you have an equation for the Vs(t)=Va*u(t).

Have a look at this

http://en.wikipedia.org/wiki/RL_circuit

Hope this helps you get going

J

#### dacrazyazn

Joined Mar 30, 2009
22
alright, so i tried to follow some of the readings... and here is what i got so far..

Vs(t)=Vo(t)+L(di(t)/dt)

Transient solution
-R/L*i(t)dt=di(t)/i(t)
-R/L*t+k=ln(i)
i(t)=ke^((-R/L)*t)

Forced solution (this is where i get stuck)
Vs(t)=Vo(t)+L*(di(t)/dt)
Vo(t)=R*i (this is the voltage across R)
di(t)/dt=-R/L*Ke^((-R*t)/L) (this is the derivative of the natural solution)
therefore plugging everything back into the original
Vs(t)=R*i-R*Ke^((-R*t)/L)

Now do i set Vs(t)=Va*u(t) ? is the process good so far?

#### jasperthecat

Joined Mar 26, 2009
20
What is your Vs(t)=Va*u(t) - a sinewave, a square wave, a step function or something else that is driving the circuit.

#### dacrazyazn

Joined Mar 30, 2009
22
I do not know. that is what was given to me from the problem, if anything i think it is a step function..

#### jasperthecat

Joined Mar 26, 2009
20
Dacrazyazn

You're right Vs(t)=Va*u(t) is a step function - this is a fancy way having a battery voltage of zero at time t <= 0 and when switched on has a value of Va when t >0.

Thus your problem becomes a L in series with a R and you are measuring the Voltage (Vo) across the R .

When switch closed

Va = R*i + L *(di/dt) (from KVL)

dt = (Ldi) / (Va - Ri)

Solving this DE gives

-(R/L)*t = ln(1 - (R/Va)*i)

1 - (R/Va)*i = e ^ (-t/(L/R))

Solving for i

i = (Va/R)*( 1 - e ^ (-t/(L/R))

i = (Va/R) - (Va/R) * e ^ (-t/(L/R))

This is the solution for the current - give it the reality check

(1) at t = 0 no current flows because Va is completely opposed by the inductor's back -emf

i = (Va/R) - (Va/R) = 0 (OK so far)

(2) at t >> 0 approaching infinity ( all the back emf from the inductor has decayed )

meaning (Va/R) * e ^ (-t/(L/R)) = 0

so i = Va/R + 0 (Looks good)

Vo = R*Va/R - R*Va/R * e ^ (-t/(L/R))

Vo = Va - Va * e ^ (-t/(L/R))

Thus from my text book definition of forced component ( a voltage proportional to the forcing or driving voltage ) in this case is the first term Va

and the natural component is - Va * e ^ (-t/(L/R)) (note sign as it opposes the driving voltage).

I think that's it

J