Help Understanding zero state RL circuit *first derivative*

Thread Starter


Joined Mar 30, 2009
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?


Joined Mar 26, 2009

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

Hope this helps you get going


Thread Starter


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


Transient solution

Forced solution (this is where i get stuck)
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

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


Joined Mar 26, 2009

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)

Your Voltage Vo = i*R

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