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Okay so I solved this question which asked me to calculate the value of R such that there is no current flowing in the galvanometer branch.

(THE ANSWER IS 100 ohm)

I used the loop rule to write the equations and solve for R.

It was after I solved the question that the trouble started. Upon closer inspection I saw that the potential drop across the 100 ohm resistor was also 2 V (Like the battery on the adjacent branch). So why is no current flowing across the 2 V battery. It can be said that the battery and the 100 ohm resistor are in parallel (Equal potential drops). How is the battery different from the 100 ohm resistor ?


3 Answers 3


It might be useful to think of some limiting cases to get some intuition. For example, assume that R starts off being infinite (open circuit), then all the current would flow through the 2V battery, which would be (12-2)/500 = 0.02A. Now, as you decrease the resistance, some of the current will flow through R as well, but only what is needed to satisfy V=IR, where V=2V. When R is very large, I will be very small. Once you get to R=100 ohms, then 100*0.02=2V, so that the 0.02 A when flowing through the resistor R will exactly produce the 2V drop and thus require no current through the 2V battery. If R goes below 100 ohms, then you will again see current flowing through the 2V battery, but in the other direction. I hope this helps.

  • $\begingroup$ Why is the 2V potential difference in the opposite direction of the battery's emf enough to stop current flow. I read that the potential of 2 points must be equal to stop current flow. How can we say that the potential difference between the positive terminal and the node (after the 500 ohm resistance) are at equal potential? $\endgroup$ Nov 24, 2019 at 8:16
  • $\begingroup$ The current meter/galvanometer (G) has zero voltage drop across it. It is an ideal device. I hope that helps. $\endgroup$
    – ad2004
    Nov 24, 2019 at 8:25
  • $\begingroup$ The galvanometer branch has a potential drop of 0 since the 12 V battery imposes a 2 V drop across it and the 2 V battery does the same but in the opposite direction. Is that what you mean to say? $\endgroup$ Nov 24, 2019 at 8:29
  • $\begingroup$ Actually, the galvanometer will always have a potential drop of zero under all conditions because it is ideal - meaning zero resistance. This is the definition of a perfect galvanometer. In the case of your problem, the node after the 500 ohm will always be at 2V potential because the 2V battery is attached there through the "ideal" galvanometer. The currents will then be whatever is needed to satisfy Ohm's law, as earlier explained. I hope this helps. $\endgroup$
    – ad2004
    Nov 24, 2019 at 8:36
  • $\begingroup$ When you say the node will be at 2V what is your reference point in the circuit? $\endgroup$ Nov 24, 2019 at 8:41

Think of a water pipe system.

  • The resistor is like a filter. It allows some water through (from any direction) but slows it down.
  • The battery is like a pump. It literally forces water from one side to the other, and allows no water to flow the opposite way.

A battery is a fairly complex electrochemical device, but it does indeed act as a one-way charge mover, which via high chemical potentials separate electrons from atoms at one terminal and reunites them at the other. This causes a constant surplus of negative charge at one terminal and a lack at the other, and this charge difference corresponds to an electrical potential energy difference, measured as voltage. Think of voltage as water pressure difference.

Those surplus electrons want to move, and the only available path they have is through the circuit (and thus they constitute a current - think of current as the water flow) because they aren't able to move through the liquid interior of the battery. This liquid (or maybe in the future non-liquid) electrolyte, as it is called, is an ion conductor, but not an electron conductor.

Sure, rechargeable batteries (accumulators) are not "one-way streets", so the analogy is not perfect and depends a lot on the exact conditions. But the one-way pump idea works fairly well in most cases.

  • $\begingroup$ “but it does indeed act as a one-way charge mover” you might want to say this applies to non rechargeable batteries $\endgroup$
    – Bob D
    Nov 24, 2019 at 11:20

The 12v battery is supplying current to R through the 500 ohm resistor that is equal to the current R would draw from the 2v battery, so they equal out.

  • $\begingroup$ Check my comment on ad2004's answer. $\endgroup$ Nov 24, 2019 at 8:19
  • $\begingroup$ "they equal out" and therefore get cancelled out by superposition? $\endgroup$ Nov 24, 2019 at 8:31
  • $\begingroup$ If you removed the 2v battery the circuit would work the same. $\endgroup$ Nov 24, 2019 at 8:37
  • $\begingroup$ If I remove the 2V battery wouldn't the 100 ohm resistor be shorted out? $\endgroup$ Nov 24, 2019 at 8:41
  • $\begingroup$ When you say that the currents equal out, how does that imply that the current in the galvanometer branch is 0 ? $\endgroup$ Nov 24, 2019 at 8:43

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