Power absorbed by multiple elements

[DODGEVIPER13]

Homework Statement

Calculate the power absorbed at t=0,t=0+,and t=200 mS by each of the elements in the circuit of fig 11.25 if us is equal to (a) -10u(-t) V (b) 20 + 5u(-t) V

Homework Equations

The Attempt at a Solution

http://www.chegg.com/homework-help/calculate-power-absorbed-t-0-t-0-t-200-ms-elements-circuit-f-chapter-11-problem-3e-solution-9780073529578-exc

The link posted above is the solution I'm trying to understand it I can also upload my work but it is more or less a copy of the solution so I saw no point. What I don't understand starting on the first part is how Vs(t)=-10u(-t) V=-10 at negative infinity to 0? First off what is u(-t) is this reactance or impedance? Furthermore if I took the limit if this as t approaches negative infinity would the function go to infinity not -10? So obviously it's not infinity even though it seems that it would, so I would have to assume u(-t)=1 for it to work. Why one because I am assuming that it has to do with current not flowing before time t=0? This also then begs another question as to why from 0 to infinity it is 0 implying that the voltage stops after time t=0 because u(-t)=0.Well anyways this is just two question of many that I have, so I am going to ask this question first and slowly propagate through my errors in this problem?

 

[rude man]

First off, I call the resistor R and the inductor L.

v_s(t) is voltage.

v_s = -10u(-t) means v_s was -10 Volts from time = -∞ to t = 0, then v_s = 0V. So at t=0 what is the current thru R? And 200 ms later what is it?

PS - L does not 'absorb' power, it stores it and discharges it. So the only element you have to worry about is R.

 

[DODGEVIPER13]

Hmm could it be i(t)=Vs/R(1-e^(-Rt/L))u(-t) which gives me -10 V at t=0 but I fear that this isn't even close?

 

[rude man]

Hmm could it be i(t)=Vs/R(1-e^(-Rt/L))u(-t) which gives me -10 V at t=0 but I fear that this isn't even close?

It's not only close, it's exactly right.

But you did not need to solve for V(t). Any function of time, like a voltage V(t), when multiplied by U(-t) means the function is V(t) until t = 0, then it's zero. It's the opposite of V(t)*U(t) which is zero until t = 0, then = 1.

Your function is V(t)*U(-t).

U(t) = 0 for t < 0, then = 1
U(-t) = 1 for t < 0, then = 0.
More generally, U(t-τ) = 0 for t < τ, then = 1
U[-(t-τ)] = 1 for t < -τ, then = 0

 

[DODGEVIPER13]

Ok so thanks for the help on the first part. Let me ask you another question is the solution posted correct? Also at t=0- the voltage across the inductor voltage is Vl(0-)=L(di(0-)/dt) then L(d(-10)/dt)=0 why is this is this just how an inductor works?

 

[DODGEVIPER13]

Oh wait the solution can't be correct because no power is absorbed by an inductor

 

[rude man]

Oh wait the solution can't be correct because no power is absorbed by an inductor

That's right.

With V(t) = 20 + 5U(-t) that means the input votage is 25V until t = 0-, the it's 20V for t >= 0+.

So you need to establish the initial current i(t) at t = 0 and then solve your equation for the current after that. Same for part (a). Then power dissipated in R = [i(t)^2]*R.

To sum up, (1) find your initial current based on the input voltage at t = 0-
(2) calculate the total current at t = 0 and t = 200 ms so you need i(t), t >= 0+
(3) compute (i^2)*R in all four cases.

When calculating the total current, remember that it consists of two parts:
1. the time-varying current due to the initial current i(0), which is the current with zero input voltage, and
2. the time-varying current due to the input voltage as though the initial current were zero.

Using the principle of superposition you add both parts to determine the current as a function of time t > 0. The two currents have different equations BTW.

 

[DODGEVIPER13]

So initial current at t=0- the current is -10 A. Which if I plug into the power equation i^2R I get 100W at t=0-.

 

[DODGEVIPER13]

Which is for R

 

[rude man]

So initial current at t=0- the current is -10 A. Which if I plug into the power equation i^2R I get 100W at t=0-.

That's correct. And BTW the current is the same at t = 0+ as at t = 0-. That's because current thru an inductor can't change instantaneously.

Now the harder part: how do you compute the power at t = 200 ms?

 

[DODGEVIPER13]

Ok so then the power at P(0+)=100 W and so is P(0). So I found an equation from the solution that I think might work so P(200 mS)=100e^(-200/250) W = 44.93 W I am goons keep looking for equations because I feel this is incorrect

 

[rude man]

Ok so then the power at P(0+)=100 W and so is P(0). So I found an equation from the solution that I think might work so P(200 mS)=100e^(-200/250) W = 44.93 W I am goons keep looking for equations because I feel this is incorrect

'Fraid so. Remember I said you need to consider two separate currents: assuming zero volts input for all t > 0, the current at t=0 and how it behaves for t>0; and the current due to the finite input voltage at t = 0+ assuming no initial current. Then you add the two currents.

Start with figuring the current for t>0 as if Vin = 0 for all t >= 0+.

 

[DODGEVIPER13]

Wait if the input voltage is zero wouldn't that make the current at those values assuming no voltage 0 too?

 

[rude man]

Wait if the input voltage is zero wouldn't that make the current at those values assuming no voltage 0 too?

The current was not zero at t=0 even if Vin = 0 for t =>0. There is current in the inductor even if you short the input at t=0.

 

[DODGEVIPER13]

Well the voltage across the inductor would be -10e^(0)=-10 V so Il=-10 A at t=0

 

[rude man]

Well the voltage across the inductor would be -10e^(0)=-10 V so Il=-10 A at t=0

No, the voltage across the inductor at t = 0- is zero. For an inductor, V = L di/dt and di/dt = 0 at t = 0- since by then the current is a nice constant -10A. At t = 0- the entire input voltage is across the resistor.

The current has had infinite time (from -∞ to 0-) to settle out.

You need to solve for the current where R is in series with L, the combo is shorted out, but an initial current runs thru R and L. What is the time response of that current?

 

[DODGEVIPER13]

L/R so 250e-3/1= 250e-3 sec

 

[DODGEVIPER13]

I looked up an equation for series RL circuits don't know if its helpful but I(t)=I(1-e^(-t/Tao)) which when I solve
-10(1-e^(-200/250)) which gives -5.506

 

[DODGEVIPER13]

Dang I guess you won't be getting back with me tonight oh well I have to turn it in at 9 tomorrow. Still I want to know about it

 

[rude man]

L/R so 250e-3/1= 250e-3 sec

?

I mean, what is the expression for i(t)?

 

[DODGEVIPER13]

Oh I was solving for the time constant. So I'm looking for an expression at t >= 0 for i(t) is it still V/R(1-e^(-Rt/L))

 

[rude man]

I looked up an equation for series RL circuits don't know if its helpful but I(t)=I(1-e^(-t/Tao)) which when I solve
-10(1-e^(-200/250)) which gives -5.506

That's the equation for the second part of the current, i.e. a step voltage is applied to an R-L series circuit with zero initial current. You actually have the correct equation for that second current component, and the correct answer for the second component of current at t = 200 ms.

But you need to add another term to account for the current that was flowing in the circuit before t=0+. This is the 1st term I spoke of.

It would probably help if you told me your academic background. The right way to explain this is to write and solve the differential equation governing the current as a function of time.

 

[rude man]

Oh I was solving for the time constant. So I'm looking for an expression at t >= 0 for i(t) is it still V/R(1-e^(-Rt/L))

That's one term. There is another, associated with the initial current at t = 0.

Think of just a series R-L circuit, shorted to itself, with the current = i₀ at t = 0. You should be able to guess what the current is as a function of time after t = 0. Hint: it's something like your expression above but obviously that expression gives i = 0 at t = 0 which you know is wrong. The same expression would also give you the current as V/R at t = ∞ and you know it should be zero instead of V/R. So take a guess at what the other current term is.

 

[DODGEVIPER13]

Hmmm could it be I(t)= Ie^(-Rt/L)

 

[rude man]

Hmmm could it be I(t)= Ie^(-Rt/L)

Sure could!
Now, looking at part (a), what is I? And how about adding in your previous equation (post 21)?

 

[DODGEVIPER13]

ok so I(t)= -10e^(-.8) - 10(1-e^(-.8))= -10 A hmmm that doesn't seem right should the V/R part be 0 as there would be no voltage across the inductor which is the only part active after 0?

 

[rude man]

ok so I(t)= -10e^(-.8) - 10(1-e^(-.8))= -10 A hmmm that doesn't seem right should the V/R part be 0 as there would be no voltage across the inductor which is the only part active after 0?

Yes and no.

Yes, V/R = 0 for part (a).
No, it's not because the voltage across the inductor is zero. It's because V = 0 for t >= 0+.

If i = -10exp(-Rt/L) and voltage across an inductor is L di/dt, how can the voltage across the inductor be zero (except at t = ∞)?

So now you can do part (b). What is the exp(-Rt/L) coefficient and what is the [1 - exp(-Rt/L)] coefficient?

 

[DODGEVIPER13]

The coefficient for [1-exp(-Rt/L)] would be 20 and 25 for exp(-Rt/L) I think?

 

[DODGEVIPER13]

Also I get I(0-)=i(0)=25 A and i(0+)= 20A

 

[rude man]

The coefficient for [1-exp(-Rt/L)] would be 20 and 25 for exp(-Rt/L) I think?

You got it! (We're doing part b now). So we have i(t) = 25exp(-Rt/L) + 20[1-exp(-Rt/L)].

Now, what is power dissipated in the resistor?