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Electronics - Determine voltage across current source

  • Thread starter turpy
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  • #1
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Homework Statement


Determine the voltage across the 2A current source
2m6lq37.png


Homework Equations


kirchhoff voltage law (kvl)
kirchhoff current law (kcl)
ohm's law

The Attempt at a Solution



I'm new to the concept of a current source. I'm not sure how there can be voltage drop across the current source, so maybe the question is asking for the voltage drop across the 10 ohm resistor that's in series with it?
Are all resistors that are parallel with the power source going to have a voltage drop equal to the source voltage? So the voltage across the 5 ohm resistor and the current source is 50V? I doubt the answer could be that simple though
I think I'm really missing something to understand how to do this question.

If I need to do KCL, it would have to be at the first junction the 2A flows into right?
and for KVL, it would be the loop with the 50V power source? Having both equations is leading me nowhere

Thanks!
 

Answers and Replies

  • #2
gneill
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An ideal current source will produce ANY amount of voltage required to maintain its specified current. So never open circuit an ideal current source :eek:

An ideal voltage source will produce ANY amount of current required to maintain its specified potential difference. So never short circuit an ideal voltage source :eek:

You'll need to find the total potential across the branch with the current source in it. Whatever the drop is across the 10Ω resistor, the remainder must be due to the current source.
 
  • #3
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An ideal current source will produce ANY amount of voltage required to maintain its specified current. So never open circuit an ideal current source :eek:

An ideal voltage source will produce ANY amount of current required to maintain its specified potential difference. So never short circuit an ideal voltage source :eek:

You'll need to find the total potential across the branch with the current source in it. Whatever the drop is across the 10Ω resistor, the remainder must be due to the current source.
Oh thank you! I didn't realize a current source was producing voltage to maintain a specific current.
(subscripts for resistors using their resistances)
Using KCL for the junction the 2A flows into:
2 + i_1 - i_5 = 0
KCL at junction below the 5 ohm resistor
i_5 - i_1 - i_10 = 0
i_10 = i_5 - i_1 = 2A
so V_10 = i_10 * R = (2)(20) = 20V

So for 50 V across the current source and 10 ohm resistor, if the voltage drop across the resistor is 20V, the voltage produced by the current source is 30V ?
 
  • #4
gneill
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Oh thank you! I didn't realize a current source was producing voltage to maintain a specific current.
(subscripts for resistors using their resistances)
Using KCL for the junction the 2A flows into:
2 + i_1 - i_5 = 0
KCL at junction below the 5 ohm resistor
i_5 - i_1 - i_10 = 0
i_10 = i_5 - i_1 = 2A
so V_10 = i_10 * R = (2)(20) = 20V

So for 50 V across the current source and 10 ohm resistor, if the voltage drop across the resistor is 20V, the voltage produced by the current source is 30V ?
Oooh, so close! Check the polarity of the drop across the 10Ω resistor.
 
  • #5
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Oooh, so close! Check the polarity of the drop across the 10Ω resistor.
So it's 70V? woah I couldn't imagine the voltage across anything in this circuit to be greater than the source voltage. But I guess it's possible since the current source produces voltage to keep the 2A constantly flowing?!
 
  • #6
gneill
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So it's 70V? woah I couldn't imagine the voltage across anything in this circuit to be greater than the source voltage. But I guess it's possible since the current source produces voltage to keep the 2A constantly flowing?!
No, there's something amiss still. Your total potential across the branch doesn't look right.

If you take the junction at the bottom of the circuit as the reference node, and if you let the voltage at the top junction (node) be Va, then you can rewrite your first KCL node equation as:

##\frac{Va - 50V}{4Ω + 1Ω} + \frac{Va}{5Ω} - 2A = 0##

Solve for Va, which also happens to be the voltage across the 5Ω resistor, as well as the branch with the current source.
 
  • #7
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No, there's something amiss still. Your total potential across the branch doesn't look right.

If you take the junction at the bottom of the circuit as the reference node, and if you let the voltage at the top junction (node) be Va, then you can rewrite your first KCL node equation as:

##\frac{Va - 50V}{4Ω + 1Ω} + \frac{Va}{5Ω} - 2A = 0##

Solve for Va, which also happens to be the voltage across the 5Ω resistor, as well as the branch with the current source.
Can you explain how you got the first term? Why are the 4 ohm and 1 ohm resistors summed in the denominator (are you treating them as resistors in series?)
I just learned about node voltage analysis in class a few hours ago; is this a method similar to that?
 
  • #8
gneill
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Can you explain how you got the first term? Why are the 4 ohm and 1 ohm resistors summed in the denominator (are you treating them as resistors in series?)
Yes, they are in series in the branch of the circuit that this term describes. If the lower node is the reference point (assumed 0V), and the node in question is at potential Va, then the current that must flow out of the node at potential Va towards the reference node is given by this term in the equation --- the net potential difference divided by the total resistance in the branch.
I just learned about node voltage analysis in class a few hours ago; is this a method similar to that?
It is exactly that, yes.
 
  • #9
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Yes, they are in series in the branch of the circuit that this term describes. If the lower node is the reference point (assumed 0V), and the node in question is at potential Va, then the current that must flow out of the node at potential Va towards the reference node is given by this term in the equation --- the net potential difference divided by the total resistance in the branch.


It is exactly that, yes.
I get it now! I just wonder why the professor assigned this question before node voltage analysis was covered.

Thank you so much for your help!
 

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