Wheatstone's bridge is a very simple but powerfull concept which has made many calculations easy. If the assumption is made that current does not flow through the galvanometer from B to C or C to B,
`V_(A)-V_(B)=V_(A)-V_(D)` `V_(B)-V_(D)=V_(C)-V_(D)`

The current in R1, has to flow through BD and that in `R_(3)` through CD. This idea has been extended to many symmetrical circuits.
If the position of the galvanometer is exchanged with that of the battery in the Wheatstone's bridge shown in the passage.

Options

A. There will be no balancing of the galvanometer
B. there will be no current in the galvanometer
C. there will be balancing if the resistances are also exchanged
D. none of these

Answer

B