How can we calculate the internal resistance of a wheatstone brigde given that Resistance 1=Resistance 2=Resistance 3=Resistance 4=1000 ohms and a galvanometer can detect as low as 0.1 milliamperes?
Calculating the internal resistance across the galvanometer
R1 and R2 are in parallel, so their effective resistance is
R3 and R4 are also in parallel. Their effective resistance is
Internal resistance is 1000ohms
R12 is now in series with R13 to give the value of the internal resistance