[Intro Circuits] Simple problem, finding initial

[jojo13]

Homework Statement

Two independent sources are given. They are active at different times.

Find the conditions at t = 0- DCSS (aka t < 0 DCSS) and t = ∞ DCSS for I_L and V_c

Homework Equations

At DCSS, replace inductor with short circuit
At DCSS, replace capacitor with open circuit

The Attempt at a Solution

So I need to find the initial and final conditions.

Let's start with initial conditions at t = 0- DCSS. The [2 + 5u(t)]V becomes 2 V. The 4u(t) I am not 100% sure about, but I think it becomes a short circuit. Then I replace the inductor and capacitor with a short and open circuit respectively because it is in DCSS.

Is this correct? The switches at 2 + 5u(t) and 4u(t) are giving me a bit of a problem understanding and I'm not sure if I'm doing it right.

As for the final conditions at t = ∞, I have no clue where to start.

 

[gneill]

I think you're almost there. Before the "t = 0" event where the 5u(t) and 4u(t) sources switch on, the leftmost voltage source will have been producing 2 V since time began, and the rightmost source will have been producing 0 V (your short circuit) for all that time. So you have to find the steady state conditions for (let's call them) V1 = 2 V and V2 = 0 V.

Do the open capacitor / short inductor analysis and find the voltages and currents.

When t = 0 arrives V1 suddenly becomes 2 V + 5 V = 7 V. V2 becomes 4 V. They stay that way from then on, so to the steady state analysis again with these new sources in place in order to find the conditions at t → ∞.

 

[jojo13]

I think you're almost there. Before the "t = 0" event where the 5u(t) and 4u(t) sources switch on, the leftmost voltage source will have been producing 2 V since time began, and the rightmost source will have been producing 0 V (your short circuit) for all that time. So you have to find the steady state conditions for (let's call them) V1 = 2 V and V2 = 0 V.

Do the open capacitor / short inductor analysis and find the voltages and currents.

When t = 0 arrives V1 suddenly becomes 2 V + 5 V = 7 V. V2 becomes 4 V. They stay that way from then on, so to the steady state analysis again with these new sources in place in order to find the conditions at t → ∞.

Ok thanks, so for the t = 0- DCSS, the 2+5u(t) becomes 2v and the 4u(t) becomes a short.

Now for the t = ∞, the 2+5u(t) becomes 7V and the 4u(t) becomes 4V? I was thinking that the 2+5u(t) becomes 2V and the 4u(t) becomes 4V

 

[gneill]

u(t) is a unit step function. It's zero for all times before t=0, then 1 for all times t > 0.

So once the u(t)'s have turned on, they stay on.

 

[jojo13]

u(t) is a unit step function. It's zero for all times before t=0, then 1 for all times t > 0.

So once the u(t)'s have turned on, they stay on.

Ahhh I see.

So just to confirm... At both t = 0 and t = ∞, the left side is 7V and the right side is 4V.

That makes a lot more sense than what I was thinking.

 

[gneill]

Ahhh I see.

So just to confirm... At both t = 0 and t = ∞, the left side is 7V and the right side is 4V.

That makes a lot more sense than what I was thinking.

Yup. To be specific, for all time up to and including t = 0^- the left side is 2 V and the right is 0 V. At t = 0⁺ the left side becomes 7 V and the right side 4 V and they stay that way indefinitely.

 

[jojo13]

Yup. To be specific, for all time up to and including t = 0^- the left side is 2 V and the right is 0 V. At t = 0⁺ the left side becomes 7 V and the right side 4 V and they stay that way indefinitely.

ok I think I got the final answers. Mind checking if I did them right. I was a bit confused on the open circuit at V_C, so I wanted to make sure it's correct.

At DCSS, replace inductor with short circuit
At DCSS, replace capacitor with open circuit For t = 0- DCSS, I got I_L = 0A and V_C = 2V

For t = ∞ DCSS, I got I_L = -4A and V_C = 11V

 

[gneill]

Looks good except for your Vc at t = ∞. The top end of the resistor should be at 4 V, the left end of the cap at 7V. I don't see an 11 V difference there...

 

[jojo13]

Looks good except for your Vc at t = ∞. The top end of the resistor should be at 4 V, the left end of the cap at 7V. I don't see an 11 V difference there...

Hmm, I must have done a sign error or something, this is the work I did for it

I first found I_L which was -4 A. Then I did KVL on the left side which I found to be -V_C + 7V - V₁ = 0. I subsituted V₁ to be (-4 A)*(1 ohm) and solved for the rest. Ended up with 11V.

 

[donpacino]

Hmm, I must have done a sign error or something, this is the work I did for it

I first found I_L which was -4 A. Then I did KVL on the left side which I found to be -V_C + 7V - V₁ = 0. I subsituted V₁ to be (-4 A)*(1 ohm) and solved for the rest. Ended up with 11V.

-V_C + 7V - V₁ = 0.

that's your problem. take another look at that equation

 

[gneill]

Look at the circuit. The capacitor is an open so the only current path is for the 4V supply through the resistor. As you say, IL is -4 A, so it's flowing down through the resistor leaving +4 V at its top end. That's the right side of the cap. The left side of the cap is fixed at +7V.

 

[jojo13]

-V_C + 7V - V₁ = 0.

that's your problem. take another look at that equation

Look at the circuit. The capacitor is an open so the only current path is for the 4V supply through the resistor. As you say, IL is -4 A, so it's flowing down through the resistor leaving +4 V at its top end. That's the right side of the cap. The left side of the cap is fixed at +7V.

-V_C + 7V + V₁ = 0.

V_C = 3 V