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Determine currents in resistors help

  1. Jul 31, 2009 #1
    Iv started physics electronics literacy at university as a requirement for my degree, i didnt do science at school and im very confused with whats going on..How to a determine a current in the resistors, please can someone give me a step by step on how they got to the answer...i have uploaded a picture of the circuit i was given

    all help is very much appreciaited, im writing a test on this stuff soon.. i know how to get Req of parrallel resister, but how do i get the currents, specifically of the 3.3k and 560 resistors?
     

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  3. Jul 31, 2009 #2

    jmb

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    If you already know how to calculate the effective resistance of parallel and series circuits, then you should be able to use Ohm's Law to calculate the total current through the circuit.

    In addition if you use the following two pieces of information: (1) the current along a series circuit is the same everywhere and (2) the total voltage drop across two pieces of circuit that are joined in parallel is the same for each piece, then you should be able to work out the things you need (btw these two rules come from the more general set of rules known as Kirchoff's laws).

    Hint: Start out by treating the entire circuit as one single effective resistor and successively break it into smaller pieces (some of which will be series circuits and some will be parallel circuits).

    Have a go at it and post your attempt back here and we will give you feedback/corrections...
     
  4. Jul 31, 2009 #3


    Thanks for your reply, ok so i worked out the total resistance to be 2909.5, then worked out the current passing through I2 (5/2909.5) and got 1.72mA which gave me a voltage of 3.44V, is that correct?

    Now what do i do from here, how would i work out the current passing through through the 1k resistor? please show me how and ill then use that knowledge to try calculate the current in the 3 on my own, i just not sure what to do now
     
  5. Jul 31, 2009 #4

    jmb

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    That's correct.
    This is the total current in the whole circuit. If by I2 you meant the 2k resistor then, yes, this is also the current going through the 2k resistor.
    Again, assuming I2 means the 2k resistor, then yes that is the correct voltage drop across that resistor.

    Since that is the voltage drop across the resistor, the voltage across the rest of the circuit (i.e. if you were to connect a voltmeter between the point just after the 2k resistor and the ground terminal of the battery) is 5V - 3.44V = 1.56V.

    Using the second rule I gave you (the voltage drop across two parts of a parallel circuit is the same for each part) this tells you that the voltage across the 3300 Ohm resistor is 1.56V (from which you should be able to work out the current in it directly from Ohm's law) and the voltage drop across the remaining subcircuit (consisting of the 1000, 560 and 470 Ohm resistors) is also 1.56V.

    Since you know the voltage drop across the subcircuit you should now be able to treat it on its own as an isolated system. To find the currents in each of its resistors just go through exactly the same process as you did for the original circuit (since it exactly resembles the original circuit but with the subcircuit replaced by a single resistor).

    Have a go at finishing it off and post your results here and then we'll check it for you...
     
  6. Aug 1, 2009 #5
    I have tried am im just not sure how to work them out :( im really confused, because im not sure which values to use for the 1k, 560 and 470 ohm resistors, i know that 560 and 470 must have the same current, but how do i work out the 1k one (because its in parrallel with the 3.3k. Really dono how to get them, should the voltage be of the resistors be calculated first or the current first? thank you for all the feedback
     
  7. Aug 3, 2009 #6

    jmb

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    No they don't! Remember: only elements in serial will necessarily have the same current going through them.

    You've already worked out the voltage drop across what I've called the "subcircuit" (consisting of the 1k, 560 and 470 Ohm resistors) so you can then ignore everything else outside that subcircuit (e.g. the 3.3k resistor) provided you use this calculated voltage drop.

    It depends, sometimes it will be easier to get the voltage first, and sometimes the current. The point is to be systematic. Let's recap on what has already been done:

    • We started by treating the entire circuit as a single component. Using the laws for adding parallel and series resistors you worked out the total effective resistance for the circuit.
    • Since you already knew the total voltage across the circuit (given by the battery) you could combine this with the calculated total resistance to find the total current in the circuit (Ohm's Law).
    • Since the current won't change until there's a junction (because of the first rule I gave you: the current is the same across series parts of the circuit) we knew that the current going through the 2k resistor was equal to this total current. So this time we could use Ohm's law to find the voltage drop across the 2k resistor. This in turn told us how much voltage was 'left over' to drive those parts of the circuit that come after the 2k resistor.
    • I told you (the 2nd rule I gave you) that the voltage drop is the same across parts of a parallel circuit, so that we knew the voltage across the 3.3k resistor was the 'left over' calculated above. Since we know resistance and voltage, this should be enough to let you calculate the current going through it using Ohm's Law again.
    • This 'left over' voltage is also the voltage drop across the 'subcircuit' (consisting of the 1k 560 and 470 Ohm resistors).

    As I said, the remaining subcircuit can be solved by repeating the same process again. Try and think what this is for yourself and then check it against the below:
    • Calculate the total effective resistance for the subcircuit using the parallel and series resistor rules.
    • Use this, the already calculated voltage drop across the subcircuit and Ohm's Law to calculate the total current in the subcircuit.
    • This total subcircuit current must also be the current going through the 1k resistor (first rule I gave you), thus you can use Ohm's Law to cacluate the voltage drop across the 1k resistor.
    • You then know the 'left over' voltage is the voltage drop across the parallel 560 and 470 Ohm resistors. Since the voltage drop across both will be the same (2nd rule I gave you) you can again just use Ohm's Law to calculate the current in each.
    Have a go at doing this and see what numbers you get. You should see that current is conserved at junctions (i.e. the outgoing current always splits into amounts that sum up to the original input current at that junction).

    If you still have difficulty then there are some additional rules I can give you. However, these rules are the result of applying the rules I've already given you to the general case. So if you can do things without them the first few times you'll end up with a much better understanding of how circuits work...

    Let us know how you get on!
     
  8. Aug 4, 2009 #7
    ok thank you so mch i think i got it...the current in the 1k is 1.24V, leaving 0.32V in the 560 and 470 ohm resistors..therefore current in the 570 ohm is 571 micro amps and the 470 is 680 micro amps
     
  9. Aug 5, 2009 #8

    jmb

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    Good that is correct, but be careful with your rounding. If you want to quote the current in the 560 Ohm resistor to 3 figures then you need to keep your intermediate calculations more accurately (the answer should be 568 microAmps but you have a little rounding error in your own calculations, similarly the 470 Ohm resistor has 677 microAmps).

    As I mentioned in my last post you should notice that the current is conserved. In fact you could have used this fact to save a little time in your calculations.

    However I would suggest you keep using the simple approach we went through here until you feel happy with how things work --- this should give you a good feeling for circuits.

    Once you are comfortable with the methods I would advise you to do this:

    Consider a simple series circuit consisting of just two resistors of resistance A and B under a total voltage V (this is often called a potential divider). Use the rules we've been using to calculate an expression for the voltage drop across the first resistor (A) and hence the remaining voltage that is applied to resistor B. The resulting formula should allow you tackle these kind of problems much faster since you should then be able to immediately calculate the voltage across each resistor and so then get the current straight from Ohm's Law. The advantage (in my opinion) is that you will also by this point understand the formula rather than just be applying it blindly.
     
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