October StudentZone Quiz Solution

You may think you need to use loop equations to solve the problem, but if you look closely at the second loop you will see that the current through R1 is defined by the 2mA current source. Therefore, the voltage across R1 can be found simply by Ohms law: VR1 = IR1 × R1 = 2mA × 1kΩ = 2V.

Sure, if you wanted to calculate all the node voltages with respect to ground then you would need to use either loop equations or network reduction, but you don’t need that to answer the question as stated.

The moral of the exercise is to always look for the simplest solution first, and don’t make the problem harder than it really is.

Comments
Anonymous
    •  Analog Employees 
    over 4 years ago

    This quiz question is an interesting mental exercise but in the real world a true floating current source like this is very difficult to realize and the possible voltage across it will always still be constrained by the power supply rails. Teaching the rules of circuit analysis using these ideal sources is great for getting an understanding the basic fundamentals but very soon the real world creeps in. When an "ideal" current source is used to model an NPN transistor for example it is not actually a source of current in that it can't produce energy. It is simply a mathematical contrivance to explain how the transistor controls the flow of current from some other power source. The can sometimes be a source ( pun intended ) of confusion for the students at first.

    Aside from a complicated system of op-amps and resistors ( look at how this is done in the ADALM1000 SMU output and it does not actually "float" ) the closest thing to a floating current source I can think of is a opto isolator with a photo transistor output. The collector and emitter of the transistor can float to just about any voltage but it is still not an actual source of power so is not a true current source. The current still needs to be supplied by some voltage source connected to the transistor.

    Doug

  • over 4 years ago in reply to jbruant

    Hi Jbruant,

    Thanks for the comments.

    I'm not sure how this stuff is taught in schools today--if it is taught--but the assumption of ideal circuit elements hopefully eliminates the need for considering other more complex models.

    We were taught that an ideal current source can have any voltage across its terminals and still produce the same current. This says it doesn't really matter about the rest of the circuit, and the voltage across R1 is only defined by the current source.

    Giving this type of quiz problem right after a study of loop equations is one way to lower the class average  

    Showing the arrows for the four loops is bound to get most students off on the wrong track.

    Best Regards,

    Walt

  • over 4 years ago

    I was tempted by this response of considering only the current driven by the current generator at the beginning but was it really correct?

    I mean it seems to depend on the model you choose for generators and can you be sure that none other branches will add a contribution?

    Applying Kirchhoff's circuits laws seemed not to be realistic in a decent time...

    So I used the superposition principle: as soon as you consider another generator than the current generator close to R1, this one can be replaced by an open circuit (ideal Norton model) which demonstrates that no other generator has any contribution to the current through R1. Q.E.D

    And if you want to use a bit more realistic model (very high resistance instead of open circuit) the superposition principle still applies and you end up with a sum of 6 terms for defining the current through R1 instead of ... many equations to solve!

    Thanks for this little exercise, back to basics!

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