Solve Circuit Analysis Homework: KVL, KCL, Ohm's Law

[Learnphysics]

Homework Statement

[PLAIN]http://img121.imageshack.us/img121/7384/1037s.png

Homework Equations

KVL, KCL, Ohm's law, voltage divider

The Attempt at a Solution

The equivlent resistance of the circuit is 25ohms. Therefore the current flowing through i2 is 20volts/25ohms = 0.8A

using KCL at the top node, 3A + i1 = .8, therefore i1 =-2.2A

The answers at the back of the book however say that i1 should be -2A, and i1 should be 1A.

Where did i go wrong, or does the book have a mis-print?

 

[staticd]

your answer makes sense to me...

 

[Learnphysics]

My concern is that, there is more than 20 volts across those two resistors.

I'm very unfamiliar with how current sources work, but maybe at the top node the current source adds to the voltage of the voltage source (somehow), for a total of 25V. Although, again i have no idea.

 

[staticd]

My concern is that, there is more than 20 volts across those two resistors.

Okay. I will tell you. That makes no sense. If you put a volt meter across those two resistors in the lab, what value do you think you would get? 20V.

Even if you put the volt meter across the current source, guess what you would get? 20V.

Elements in parallel with a voltage source have the same potential as the sum of the voltage source(s).

Try doing a source transformation with the voltage source and the resistors and see what you get...

The beautiful part about electrical engineering (circuit analysis, in this case) is that there are many ways to skin a cat (but usually only one best way).

 

[skeptic2]

I concur with i1 = 0.8A and i2 = -2.2 A.

Simulated in PSPICE and it also gets 0.8 A and -2.2 A.

 

[Learnphysics]

Okay. I will tell you. That makes no sense. If you put a volt meter across those two resistors in the lab, what value do you think you would get? 20V.

Even if you put the volt meter across the current source, guess what you would get? 20V.

Elements in parallel with a voltage source have the same potential as the sum of the voltage source(s).

Try doing a source transformation with the voltage source and the resistors and see what you get...

The beautiful part about electrical engineering (circuit analysis, in this case) is that there are many ways to skin a cat (but usually only one best way).

Hmm, i read somewhere that the current source will 'change' the voltage across it's terminals in order to ENSURE that 3A goes through it. eg. it forces 3A through it.

It's put into a circuit with a 25 ohm resistance. Therefore the current source SHOULD have V=ir, 25*3 = 75V at it's terminals right?

as it's in parralell with the 20v, as you said we'd SUM the 75 and 20, to get 95.

I'v simulated it in spice and i got -2.2A and .8A... but i still don't understand why the above theory is wrong. What am i missing.

 

[Tweedle_Dee]

That would be true if the 20V supply was not there. But being an ideal voltage source, it will sink or source current in order to keep the node at 20V.

 

[Learnphysics]

That would be true if the 20V supply was not there. But being an ideal voltage source, it will sink or source current in order to keep the node at 20V.

sorry, what did you mean by sink or source the current? and which node will be kept at 20?

 

[Tweedle_Dee]

sorry, what did you mean by sink or source the current? and which node will be kept at 20?

Sink current means current going into the voltage source. Source current means current going out of the voltage source. The node I am talking about is the point where the voltage source and current source meet. A current source will force 3A and put no constraints on the voltage. Likewise, the voltage source will force 20V but put no constraints on the current. So, with this knowledge, you know for certain that 3A is coming out of the current source and 20V is across the voltage source. Now you just calculate the currents using KCL.

 

[Learnphysics]

Sink current means current going into the voltage source. Source current means current going out of the voltage source. The node I am talking about is the point where the voltage source and current source meet. A current source will force 3A and put no constraints on the voltage. Likewise, the voltage source will force 20V but put no constraints on the current. So, with this knowledge, you know for certain that 3A is coming out of the current source and 20V is across the voltage source. Now you just calculate the currents using KCL.

Ah, vey helpful.

Thanks!