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  1. papiLG is offline Junior Member Pro Subscriber
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    question on problem

    Does any one know how to solve this problem?
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    What is the magnitude of current along the center vertical wire?
    *0.75A

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  3. anonymous is offline Junior Member Pro Subscriber
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    Quote Originally Posted by papiLG View Post
    Does any one know how to solve this problem?
    Click image for larger version. 

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    What is the magnitude of current along the center vertical wire?
    *0.75A
    The real answer is obviously not correct: zero amps will flow through the center wire and all the other wires, because this is an open circuit.

    But the question gives you some current values, so we have to assume there is a complete circuit (not pictured). So, imagine a wire going from point A to a resistor RA, with the other side of RA to the center wire. Then imagine a wire going from point B to a resistor RB, with the other side of RB to the center wire. Now you can use Ohm's law with the given values to work out the rest of the drawing.

    But you don't really have to. We have a DC circuit, there are two currents flowing into each other at a junction, so the currents will add together and flow out of that junction. 1/4 A + 1/2 A = 3/4 A = 0.75 A. Another way of looking at this is that there are two batteries connected in parallel (negative joined with negative). Like any time you connect batteries in parallel, the current sourced from one will add up with the current sourced from the other. Everything else is extra information.

    For the sake of completeness:

    IB = 1/4A, EB = 8V, and the given resistor on B is 4Ω. The unknown resistor RB is:
    RB + 4Ω = 8V / 0.25A
    RB = 32 - 4 = 28 Ω

    Following the same process on the A side, we get
    IA = 1/2A, EA = 2V, and the given resistor on A is 6Ω. The unknown resistor RA is:
    RA + 6Ω = 2V / 0.5A
    RA = 4 - 6 = -2 Ω

    So check out your local electronics store and see if they have a negative two ohm resistor for sale. Obviously that can't be right. But look at that drawing again for the trick. 2V is not actually on the voltage source, it's in between the two voltage sources, which are connected backwards (against each other, or subtractively). That means EA is actually either 6V or 10V.

    IA = 1/2A, EA = 6V, the given resistor is 6Ω. The unknown resistor RA is:
    RA + 6Ω = 6V / 0.5A
    RA = 12 - 6 = 6Ω

    You can use the same process with EA = 10V and get RA = 14Ω. Either way, the real problem was figuring out that it's a trick question.

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    Quote Originally Posted by anonymous View Post
    The real answer is obviously not correct: zero amps will flow through the center wire and all the other wires, because this is an open circuit.

    But the question gives you some current values, so we have to assume there is a complete circuit (not pictured). So, imagine a wire going from point A to a resistor RA, with the other side of RA to the center wire. Then imagine a wire going from point B to a resistor RB, with the other side of RB to the center wire. Now you can use Ohm's law with the given values to work out the rest of the drawing.

    But you don't really have to. We have a DC circuit, there are two currents flowing into each other at a junction, so the currents will add together and flow out of that junction. 1/4 A + 1/2 A = 3/4 A = 0.75 A. Another way of looking at this is that there are two batteries connected in parallel (negative joined with negative). Like any time you connect batteries in parallel, the current sourced from one will add up with the current sourced from the other. Everything else is extra information.

    For the sake of completeness:

    IB = 1/4A, EB = 8V, and the given resistor on B is 4Ω. The unknown resistor RB is:
    RB + 4Ω = 8V / 0.25A
    RB = 32 - 4 = 28 Ω

    Following the same process on the A side, we get
    IA = 1/2A, EA = 2V, and the given resistor on A is 6Ω. The unknown resistor RA is:
    RA + 6Ω = 2V / 0.5A
    RA = 4 - 6 = -2 Ω

    So check out your local electronics store and see if they have a negative two ohm resistor for sale. Obviously that can't be right. But look at that drawing again for the trick. 2V is not actually on the voltage source, it's in between the two voltage sources, which are connected backwards (against each other, or subtractively). That means EA is actually either 6V or 10V.

    IA = 1/2A, EA = 6V, the given resistor is 6Ω. The unknown resistor RA is:
    RA + 6Ω = 6V / 0.5A
    RA = 12 - 6 = 6Ω

    You can use the same process with EA = 10V and get RA = 14Ω. Either way, the real problem was figuring out that it's a trick question.

    Click image for larger version. 

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ID:	530Click image for larger version. 

Name:	Screenshot_20210922-143446.png 
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ID:	531
    Thats a really cool circuit. How did you make that? The question seems like really roundabout way to express Kirchhoff's Current Law.

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  7. anonymous is offline Junior Member Pro Subscriber
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    Quote Originally Posted by RabbleRabble View Post
    Thats a really cool circuit. How did you make that? The question seems like really roundabout way to express Kirchhoff's Current Law.
    Look up EveryCircuit on your phone's app store.

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