I am having trouble understanding the working of a Wheatstone bridge. I am looking for an intuitive understanding of the process. Basically, I would like to have the following questions answered:

  1. Why do the resistors connected directly to negative terminal of the battery affect the current through the galvanometer?
  2. What exactly stops the current from splitting apart and flowing through the galvanometer?

Ok so I first have taken the diagram from the Wikipedia page for reference and put it here. enter image description here

Now if you are happy with the idea of how the potential divider works...enter image description here

… then I hope that you can see that $R_1$ and $R_2$ in the Wheatstone Bridge diagram form a potential divider and there is another potential divider with $R_3$ and $R_x$ — and the points with '$V_\rm{out}$' for these potential dividers are $\rm D$ and $\rm B$ in the Wheatstone Bridge diagram.

Now the intuitive approach is to say that if these two potential dividers have the same $V_{\rm{out}}$ then points $\rm D$ and $\rm B$ in the wheatstone Bridge diagram will be at the same potential or voltage — so the difference between them will be $0\,\rm{V}$ and no current will flow through the galvanometer. If, however, the two potentials at $\rm D$ and $B$ are different then a current will flow through the galvanometer — hope this is useful.

[Note a technical point here — when no current flows between $\rm D$ and $\rm B$ then the current through $R_1$ and the current through $R_2$ must be the same — so we can use the normal potential divider equation. Similarly for the currents through $R_3$ and $R_x$. When there is a current flowing through $V_G$ then it will be much harder to calculate exactly what is happening.]

So to answer your questions…

  1. They are connected to the negative terminal to form two potential dividers.
  2. If the potential (or voltage) at $\rm D$ and $\rm B$ is the same then no current will split off and flow because there is no potential difference driving the current through the galvanometer.

Revision of the Potential Divider

The potential divider shown abover is a fundamental circuit in electronics. Perhaps the most fundamental. (For example, If you replace $R_{top}$ or $R_{bottom}$ with a capacitor you get a high pass or a low pass filter).

The first point about the Potential divider is that the current, $I$, through $R_{top}$ and $R_{bottom}$ is the same because we assume that no current goes off at the side from the junction in between the two resistors.

Now the total resistance of the two resistors is $R_{\rm{top}} + R_{\rm{bottom}}$ and the total voltage (or potential) applied is $V_{\rm{in}}$ so by Ohm's law we can find the current, $I$,

$$ I = {V_{\rm{in}} \over R_{\rm{top}} + R_{\rm{bottom}} }$$

Now the voltage, $V_{\rm{out}}$ is the voltage (or potential) over only $R_{\rm{bottom}}$ so we can use Ohm's law to find this again because we know the value of the current, $I$, and the resistor $R_{\rm{bottom}}$;

$$ V_{\rm{out}} = I \times R_{\rm{bottom}} $$


$$ I ={ V_{\rm{out}} \over R_{\rm{bottom}}} $$

and eliminating $I$ we get

$$ {V_{\rm{in}} \over R_{\rm{top}} + R_{\rm{bottom}} } ={ V_{\rm{out}} \over R_{\rm{bottom}}} $$


$$ { V_{\rm{out}} \over V_{\rm{in}} } ={ R_{\rm{bottom}} \over R_{\rm{top}} + R_{\rm{bottom}} }$$

and so

$$ { V_{\rm{out}} } = V_{\rm{in}} \times { R_{\rm{bottom}} \over R_{\rm{top}} + R_{\rm{bottom}} }$$

This is the equation for the potential divider. Note that the fraction on the right hand side ${ R_{\rm{bottom}} / (R_{\rm{top}} + R_{\rm{bottom}} })$ is a number between $0$ and $1$ so the output voltage is smaller than the input voltage or we could say the 'voltage is divided' or the 'potential is divided' by the 'potential divider'.

  • $\begingroup$ Sorry I'm not familiar with a potential divider $\endgroup$ – Apoorv Apr 13 '15 at 15:49
  • $\begingroup$ are you familiar with Ohm's law? $V=IR$? - I will add something about the potential divider as well. $\endgroup$ – tom Apr 13 '15 at 15:54
  • $\begingroup$ Yes, I am all good with Ohm's law $\endgroup$ – Apoorv Apr 13 '15 at 15:57
  • $\begingroup$ @Apoorv - great, thanks for marking as solution, hope this was helpful - I am going to edit the title to mention the potential divider because there is not anything I can see on this site yet about the potential divider.. $\endgroup$ – tom Apr 13 '15 at 16:20

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