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Find the Current at a Certain Point

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    For the circuit in the diagram below, R1 = 34.0 Ω, R2 = 20.0 Ω, ε1 = 8.50 V, ε2 = 10.5 V and ε3 = 10.5 V are connected as shown in the figure.


    What is the value of current I1?

    2. Relevant equations
    Kirchoff's Laws

    3. The attempt at a solution

    I know i have to use Kirchoff's Law, but the day we were supposed to do them class was cancelled. This is part of an online assignment that's due tonight, and it's the last question that i just can't figure out.
    Last edited: Jan 22, 2012
  2. jcsd
  3. Jan 22, 2012 #2
    I can't see the diagram -- the site you linked to asks for a login name and password. Can you copy the diagram to imgur or something so we can see it?

    However, I am familiar with this type of problem, so I can give some pointers so maybe you can do it yourself.

    0. You probably need Ohm's Law: V = IR. That is, the voltage drop across a resistor is the current through it, times the resistance.

    And Kirchoff's Laws are:

    1. The total voltage change around a loop is zero. This is actually hard to explain in words but very easy to understand. Imagine a circuit in the shape of a square. On the west side, there is a battery, and on the north, east and south sides there are identical resistors. Now we can sum the voltage around the loop by starting in, say, the south-west corner and walking clockwise around the loop. The voltage goes up by +ε as we cross the battery. Across the north side it drops by - 1/3ε, and same for the east and south sides, so that when we get back to the south-west side, the total voltage change is 0. Now this kind of thing will happen for any loop at all, even if it contains various batteries and resistors, and, more importantly, even if the loop in question is embedded in a larger circuit. It is always true, for any loop you choose to look at.

    2. Current is conserved, so that if you have a junction where several wires send current in and several wires bring current out, the sum of the currents going into a junction is the same as the sum of the currents going out. Very intuitive, and similar to the first law, it is true for any junction.

    The tricky part is to use those three laws to write down equations involving currents and voltages across resistors, and then solve those equations. If you have a given circuit diagram and a complete list of all the battery emf's and resistor resistances, then it is always possible to find all currents and voltages at all points in the circuit. But it takes a little fiddling with the algebra before you get the answer.
  4. Jan 22, 2012 #3
    Sorry about the Picture thing,
    so, if i have two batteries in the same loop, do i add together their voltage?
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