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While studying Wheatson bridge I got a minor confusion. Two loops are present in [the bridge section of ] this circuit. My question is when two currents are flowing in opposite direction and net current is obtained in specific direction then what is happening? Can someone please explain this interaction of two currents to give one net current?

  • My second question is that when galvanometer show zero deflection then would current still flow in two loops.
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I'm assuming that by "two currents flowing in opposite direction," you are talking about the branch which containers the galvanometer. In reality, you have a net current flowing one way which is modeled in the analysis by two "loop" currents. The loop currents are strictly mathematical values which are superposed to give the value of the net current. (Even the net current is only a model value. Actual charge displacement and speed is MUCH more complicated.) (Edit per Floris) The reason you can do this superposition of branch currents is because the equations for currents and voltages are linear.

When the galvanometer shows zero, there is no current through the junction, but there are currents through the side branches/resistances. The potentials on either side of the galvanometer will be equal to each other, so no net current flows.

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  • $\begingroup$ Yes - worth mentioning that the equations are linear, so you can use superposition; this means that sometimes there are "easy" ways of breaking the problem into smaller problems (like the loops) and the final result is just the sum of all these smaller solutions. $\endgroup$
    – Floris
    Commented Aug 15, 2016 at 20:03
  • $\begingroup$ @BillN@Floris My main question is during the interaction of two opposite flowing current do they cancel each other effect or maintain their integrity $\endgroup$
    – Mahin
    Commented Aug 16, 2016 at 1:40
  • $\begingroup$ As I said, there are NOT actually two opposite flowing currents. Those are just mathematical constructs which allow you to solve the system because the equations are linear. The "real" current is dependent upon the potential difference between the two sides of the galvanometer. A zero current is a sign that the potentials are equal. $\endgroup$
    – Bill N
    Commented Aug 16, 2016 at 2:19
  • $\begingroup$ @BillN in the answer you gave you quoted that charge and speed is much more complicated so can you give me a simple idea how this is possible. $\endgroup$
    – Mahin
    Commented Aug 17, 2016 at 9:09
  • $\begingroup$ First, I said "displacement of charge." Second, I don't know what you mean by "this": a) my statement that it is complicated, (you want me to simplify the complexity?) or b) there aren't two currents. Basically, electrical current in a wire is a macroscopic phenomenon which is the summation of microscopic charge migration. Unless you're dealing with condensed matter quantum systems, and have extremely $$$$ equipment, you never measure individual charge carriers. $\endgroup$
    – Bill N
    Commented Aug 19, 2016 at 13:31

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