# Homework Help: [Intro Circuits] Simple problem, need help finding initial

1. Dec 6, 2015

### jojo13

1. The problem statement, all variables and given/known data

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 IL and Vc

2. Relevant equations

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

3. 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.

2. Dec 6, 2015

### Staff: Mentor

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 → ∞.

3. Dec 6, 2015

### jojo13

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

4. Dec 6, 2015

### Staff: Mentor

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.

5. Dec 6, 2015

### jojo13

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.

6. Dec 6, 2015

### Staff: Mentor

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.

7. Dec 6, 2015

### jojo13

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 VC, 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 IL = 0A and VC = 2V

For t = ∞ DCSS, I got IL = -4A and VC = 11V

8. Dec 6, 2015

### Staff: Mentor

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...

9. Dec 6, 2015

### jojo13

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

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

10. Dec 6, 2015

### donpacino

-VC + 7V - V1 = 0.

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

11. Dec 6, 2015

### Staff: Mentor

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.

12. Dec 6, 2015

### jojo13

-VC + 7V + V1 = 0.

VC = 3 V

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