Each resistor shown in figure has a resistance of `10(Omega)`and the battery has an emf of 6 V.Find the current supplied by the battery .

Explanation

Suppose a current I starts from the positive terminal of the battery. By symmetry,it divides equally in the resistors ab and fe, so that each of these carries a current `i/2.`The current going into the megative terminal is also I and bu symmetry , equal currents should come from ed and bc. Thus, the current in ed is also `(i/2)`and hence there will be no current in eb.
We have,
`V_(a)-V_(c)=(V_(a)-V_(b))+(V_(b)-V_(c))`
or, `6V=(i/2)xx10(Omega)+(i/2)xx10(Omega)`
giving `i=0.6A.`
This is a balanced Wheatstone bridge.