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Here I have two almost identical circuits. The only difference is that the first one has a current source with the voltage source. Shouldn't the current in the upper circuit be v/r, or 10 V/1000 Ω = 0.01 A? Why is it 5 A here?

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    \$\begingroup\$ The point between R7 and the current source is not ground. Therefore the voltage accross R7 is not 10V. See answers bellow for what it is. \$\endgroup\$ Oct 24, 2017 at 21:26
  • \$\begingroup\$ ...In order to apply the v=ir rule (a.k.a., Ohm's Law), you have to measure the right V and the right I. The I is the current through a resistor, the V is the voltage dropped by the same resistor (i.e., the voltage difference between its two end points), and the R is the resistance of the same resistor. \$\endgroup\$ Oct 24, 2017 at 21:29

4 Answers 4


You can think of a current source as a smart voltage source. It creates a voltage across itself so that the desired current is reached in the attached circuit.

So, looking at your circuit...


simulate this circuit – Schematic created using CircuitLab

Ask yourself, what voltage would need to be at point B in order to have 5A passing thorough R1

Well, from OHM's law, the voltage drop across R1 must be 5,000V, so point B must be at 10V - 5,000 = -4,990V.

So the current source is generating a large negative voltage at its top.

Another way you can look at it is, for 5A to flow through this circuit, the circuit equivalent resistance must be \$2\Omega\$. The current source then must have an effective resistance of \$-998\Omega\$. Yes that is correct, the current source has a negative resistance in this instance.

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    \$\begingroup\$ We're going to need a bigger heatsink. \$\endgroup\$
    – nekomatic
    Oct 25, 2017 at 8:32

No, the voltage across the resistor is \$V_R = I\cdot R = 5 \textrm{A} \cdot1\textrm{k} \Omega = 5000\:\textrm{V}\$

Hence, the voltage across the current source is \$10\textrm{V} - 5000\textrm{V} = -4990\textrm{V}\$

Why is that? Well because you have a current source in your circuit.

A current source is a circuit element that maintains a prescribed (set) current (5A) regardless of the voltage across its terminals.

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    \$\begingroup\$ And adding a real-world caveat to this - a real voltage source maintains a constant voltage up to some maximum current, and a real current source maintains a constant current up to some maximum voltage. In your simulator, your current source will have no problems delivering -4990V. In real life, you may find it behaves a bit differently! \$\endgroup\$
    – Graham
    Oct 24, 2017 at 16:33

Nothing can "override" a law, this is what is making it a law. In the upper circuit we know the current forced by current source to be 5A. It makes the voltage drop on the resistor to be Vr = 5A * 1K = 5000V. But we also know that the total voltage on the branch is 10V as forced by the ideal voltage source on the left. It makes the voltage drop on the current source to be 10V-5000V = -4990V. Because the ideal current source does not mandate any specific voltage drop on it, this negative drop is perfectly legitimate.

  • \$\begingroup\$ Well, reality can override it. After all, the "law" is for ideal resistors, of which I have none in my parts kit ;-) \$\endgroup\$
    – Cort Ammon
    Oct 24, 2017 at 19:18
  • \$\begingroup\$ @CortAmmon Well, every law comes packed with conditions it applies to. Ohms Law is not restricting ideal current sources. \$\endgroup\$
    – Eugene Sh.
    Oct 24, 2017 at 19:21
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    \$\begingroup\$ @CortAmmon: I'd say that Ohm's Law defines resistance. The fact that a device called a "100 ohm" resistor may have a resistance which isn't constantly 100.0000 ohms doesn't mean that it disobeys Ohm's Law; it simply means that its resistance isn't a constant 100.0000 ohms. \$\endgroup\$
    – supercat
    Oct 24, 2017 at 23:17
  • \$\begingroup\$ @CortAmmon actually OHM law really states "The resistance of an object is the amount of voltage you will measure across it if you feed one amp through it." That applies to all things. it may vary with the rate of change of the applied current and may not be linear, but at any time point and given value it always applies. \$\endgroup\$
    – Trevor_G
    Oct 25, 2017 at 12:31
  • \$\begingroup\$ @CortAmmon That means the R value is a measured value from OHM's law. The resistors in you box are marked in OHM'S law resistance.. ideal or not. They have a tolerance because the manufacturer / standard defines a "close enough" percentage range for each family. \$\endgroup\$
    – Trevor_G
    Oct 25, 2017 at 12:37

As others stated a constant current source provides a steady current even if the resistance of the load varies, which is very valuable to circuit designers.

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For example, in the above circuit the load could be a resistor or whatever. The current source supplies 50mA to it regardless of whether the load resistance varies. So in your problem 5A current goes through the 1k resistor and as the result the resistor drops 5,000V.


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