Recent content by Harsharma17

  1. H

    How does current division apply to find current in QA and QB?

    Yes, voltage, or potential difference is required for flow of current, just like two terminals of battery provide the voltage to the circuit, and electrons move from one point to another till this difference is cancelled. Here's the link for a page about voltage and current flow with various...
  2. H

    How does current division apply to find current in QA and QB?

    If there wouldn't be voltage difference between A and B, then how would current of 0.66 A(as given in the answer) would flow from A to B? If we don't consider the answer for a moment, it is clear from the figure that the circuit is not a balanced wheatstone bridge, as 3/6 is not equal to 6/3...
  3. H

    How does current division apply to find current in QA and QB?

    I tried putting H-shape wire in place of AB, but that would mean all the current would flow from left part of circuit to right part through AB, but that's not the case. Current is going from A and B to Q, not only from either A or B. But I can't draw it in any other way. I was unable to upload...
  4. H

    How does current division apply to find current in QA and QB?

    I think current division rule would give I current in 6ohm, and 2 I current in 3 ohm, because current is inversely proportional to resistance.
  5. H

    How does current division apply to find current in QA and QB?

    Providing the link to that site would be useless, because the site allows it's answers to be viewed by only those who have an account there. It is a study course-type site, and most of the questions asked are neglected there, so I asked it's explanation here. Nevertheless, here's the link...
  6. H

    How does current division apply to find current in QA and QB?

    Sorry for not providing the link and diagram, now I have edited the question to give these details.
  7. H

    How does current division apply to find current in QA and QB?

    Here's the diagram of the circuit: And here's the solution I found on a site on Google: I tried to follow through the solution given, but can't understand how it is applying current division rule to find current in QA and QB, and subtracting current values to find the current in AB is still...
  8. H

    How to solve this network of resistors for equivalent resistance?

    Yes, now I able to separate them in parallel circuits. Thanks for the reply!
  9. H

    How to solve this network of resistors for equivalent resistance?

    Thanks for the reply! Going through your suggestion, I think PA and PB are in parallel, and QA and QB are also in parallel. Solving them would lead to 2 ohm on each side of AB in series. And the equivalent resistance would be 4 ohm. I got the answer! Thanks for the help.
  10. H

    How to solve this network of resistors for equivalent resistance?