Engineering DC Steady State Circuit Analysis: Solving for iL, i, and Vc in a Simple Circuit

[jdawg]

Homework Statement

DC steady state circuit analysis: Consider the circuit shown below. The switch shown in the problem opens at time t = 0 seconds. If the circuit is in DC steady state at , find values for i_L(0^-), i_L(0⁺), i(0^-), i⁺, V_c(0^-), V_c(0⁺). There is no RL or RC analysis to be performed on this problem. On your answer for the current, a positive value indicates current is flowing in the direction shown. A negative value for current will indicate that the current is flowing opposite the direction shown.

Homework Equations

The Attempt at a Solution

Ok so I know that i_L(0^-)=i_L(0⁺) because current is constant through an inductor. And I know that V_c(0^-)=V_c(0⁺) because voltage across a conductor is constant.

I feel like I don't know some of the basic concepts I need to solve this. Is I_s(t)=0 since the switch is up? Is the 26v voltage source the same as the voltage across the 10 ohm resistor? Vc is supposed to equal 6 volts and iL and i are supposed to be 1.5 amps. I can't quite figure it out :(

 

[JBA]

First, what is Vc at the time the switch opens; and then, what part of the circuit can you remove (i.e. has no current) after the switch is opened.

 

[jdawg]

Does Vc=0 when it opens? Then you can remove the part of the circuit to the far right where Vc is?

 

[JBA]

No that is not correct, what happens when a voltage is applied to a capacitor?

 

[jdawg]

The voltage doesn't instantaneously change right? It builds up over time? Sorry, I'm not sure if that's what you were trying to get me to say.

 

[JBA]

The problem states that there is no induction effects to be considered but that is irrelevant for your problem. Once the voltage source is applied to the circuit it will charge the capacitor to the voltage source value; in this case, 26 volts, and the capacitor voltage will remain at that level once the voltage source is removed until some event provides a circuit path through which the capacitor can discharge. Knowing that the voltage source can only be effective as long as both of its poles are connected what is then the result (the part of the circuit that can be removed) once the switch is opened?

 

[jdawg]

The branch with the 10 ohm resistor?

 

[JBA]

That is correct, now with the capacitor as the voltage source (remember current flows from + to - ) and to the ground where are there going to be currents in the remaining circuit and what direction will they be?

 

[jdawg]

i(t) should be going down through the 4 ohm resistor and the current iL(t) should be going the opposite direction shown in the picture. Is there only one current now flowing through the loop with the 12 and 4 ohm resistors?

 

[JBA]

You are correct, the flow will be reversed through L, and the capacitor will discharge from + to - through both resistors the same direction as shown for the 4 ohm resistor but at different amperages.

 

[jdawg]

Ok awesome! So how do you actually start calculating the currents and Vc?

 

[gneill]

The problem states that there is no induction effects to be considered but that is irrelevant for your problem. Once the voltage source is applied to the circuit it will charge the capacitor to the voltage source value; in this case, 26 volts, and the capacitor voltage will remain at that level once the voltage source is removed until some event provides a circuit path through which the capacitor can discharge. Knowing that the voltage source can only be effective as long as both of its poles are connected what is then the result (the part of the circuit that can be removed) once the switch is opened?

No, the capacitor will not charge to 26V. There are resistors forming a voltage divider between the source and the capacitor. In particular the 10 Ω resistor in conjunction with the 12 Ω and 4 Ω resistors in parallel form a voltage divider that supplies the capacitor.

To find the steady state voltage on the capacitor (i.e. prior to the switch opening and the instant after it does) treat the capacitor as an open circuit and the inductor as a short circuit, then analyze the remaining circuit. At steady state the capacitor voltage will be the same as that across the 4 Ω resistor. The current through the inductor will be the same as that through the 4 Ω resistor. jdawg's second circuit diagram depicts the situation correctly.

Immediately after the switch opens the capacitor has the same voltage as the instant before, and the inductor the same current. The inductor's current direction will not change! It will maintain the same current in the same direction in the instant after the switch opens. The problem states that RL and RC effects are not to be considered, which means we don't need to look at the time varying voltages and currents after t = 0⁺. It does not mean we ignore the constant current engendered by the inductor or the constant voltage engendered by the capacitor at t = 0⁺.

 

[jdawg]

Ok that makes more sense. I tried doing KVL around the 12 ohm and 4 ohm resistor loop and everything just kind of blew up :( You can't use ohm's law, can you? I feel like there are too many variables... How can you find Vc without finding the current i(t) first or vice versa??

 

[gneill]

Ok that makes more sense. I tried doing KVL around the 12 ohm and 4 ohm resistor loop and everything just kind of blew up :( You can't use ohm's law, can you? I feel like there are too many variables... How can you find Vc without finding the current i(t) first or vice versa??

Personally I'd opt for nodal analysis as there's only one essential node in the circuit in question. That would yield the potential at the top of the resistors, which also happens to be V_c. Then the current through the 4 Ω resistor would be a simple matter of Ohm's Law.

 

[jdawg]

THANK YOU SO MUCH! That problem was driving me crazy!