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schematic

simulate this circuit – Schematic created using CircuitLab

From what I understand, the magnitude and direction from an independent current source is constant. What would happen in the circuit above? If currents flow they way they're supposed to from the current source, KCL is violated. So how would the circuit above behave?

How about this one?

schematic

simulate this circuit

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  • \$\begingroup\$ A real constant current source has a compliance range (limited voltage range over which it can provide the stated current) but assuming perfect ideal current sources this circuit cant work the current through the loop cant be both +10A and -1A at the same time. Any simulation package will tell you its unsolveable. \$\endgroup\$ – Warren Hill Mar 18 '14 at 16:38
  • \$\begingroup\$ @WarrenHill What about the second diagram? Is that unsolvable too? \$\endgroup\$ – dfg Mar 18 '14 at 16:40
  • \$\begingroup\$ The second diagram is solvable there is no voltage across either resistor so there is zero current flowing. As already answered by Andy. \$\endgroup\$ – Warren Hill Mar 18 '14 at 16:42
  • \$\begingroup\$ @WarrenHill Sorry, I made a mistake in the second diagram. I corrected it - how would it behave now? \$\endgroup\$ – dfg Mar 18 '14 at 16:48
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In circuit 1 you get infinite voltages and in circuit 2 you get zero current.

EDIT due to wrong circuit posted in question - circuit 2 takes 45mA and I'm not inclined to feed the reasons on a plate to the OP because this sort of stuff should be reasoned through.

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  • \$\begingroup\$ Could you explain why? And how this doesn't result in a contradiction? \$\endgroup\$ – dfg Mar 18 '14 at 16:39
  • \$\begingroup\$ Also, I made a mistake in circuit 2, that I have now fixed. \$\endgroup\$ – dfg Mar 18 '14 at 16:49
  • \$\begingroup\$ I've amended my answer but you need to show some reasoning skills to figure this out rather than lead you through the stages by hand. I've provided answers now please try and rationalize this - for your own good dude. \$\endgroup\$ – Andy aka Mar 18 '14 at 17:06
  • \$\begingroup\$ Thanks. Using Ohm's Law, wouldn't the current be 90 mA? (since the voltage difference is 9V and the resistance is 100 ohms). Also the current would be flowing the "wrong" way into the first battery (from the negative to positive terminal). Wouldn't this somehow affect the voltage of the battery and the behaviour of the circuit?? \$\endgroup\$ – dfg Mar 18 '14 at 17:10
  • \$\begingroup\$ What resistance do you have? These are theoretical voltage sources and not batteries (although batteries would behave somewhat similarly). \$\endgroup\$ – Andy aka Mar 18 '14 at 17:27
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In circuit 2, each resistors pin has the same voltage (V1 = V2 = 1V). Since there is no voltage difference between these resistors pins, there is also no current flowing through them.

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  • \$\begingroup\$ Sorry, I made a mistake in the diagram that has now been fixed. \$\endgroup\$ – dfg Mar 18 '14 at 16:49
  • \$\begingroup\$ Also what's the current in the resistors? Also can you explain how you got your answers? \$\endgroup\$ – dfg Mar 18 '14 at 17:01
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As stated in an earlier comment the first circuit is not solvable because its a simple series circuit and therefore must have the same current everywhere It can't be both +10A and -1A.

Now we come to an analysis of the second circuit.

It is also a series circuit so must have the same current everywhere.

From Kirchhoff's voltage law we know that if you follow any closed loop all the way round the voltages must sum to zero.

$$10 \text{ volt} + I \cdot 100 \ \Omega - 1 \text{volt} + I \cdot 100 \ \Omega = 0 \text{ volt}$$

$$9 \text{ volt} + I \cdot 200 \ \Omega = 0 \text{ volt} \Rightarrow I = - \frac{9}{200} = - 45 \text{ mA}$$

The minus sign just implies that the current is in the opposite direction to the one I originally guessed.

Working from the bottom of the diagram up the bottom node is 0V the voltage at the top of the bottom resistor is \$ 0.045 \text{ A} \cdot 100 \ \Omega = 4.5 \text{ volt}\$

The voltage at the bottom of the top resistor is \$ 4.5 \text{ volt} + 1 \text{ volt} = 5.5 \text{volt}\$ with 10 volt at the top node with a current of 45mA circulating counter clockwise around the loop.

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I will just boil this down to a couple of fundamentals:

  • two-terminal elements placed in series have equal current through them.
  • an ideal current source has exactly its specified current through it.

You can see that your first circuit creates a contradiction between these two rules, because you have ideal current sources in series and their specified currents are different. This means there is something wrong with your model. Either there are alternate paths for current to travel that aren't shown in the model, or the current sources are not ideal.

The dual of those rules are:

  • two-terminal elements placed in parallel have equal voltage across them
  • an ideal voltage source has exactly its specified volage across it

An ideal voltage source might have any current through it that's required to be consistent with the specified voltage.

Your second circuit doesn't create any contradiction because the voltage sources aren't placed in parallel. The net result will just be a current flowing in to the positive terminal of V1, which is no problem for an ideal voltage source (but might be a problem for some kinds of real voltage sources).

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