Question

How can we calculate the internal resistance, voltage out put due to unbalanced condition, current through galvanometer for unbalanced and smallest charge in resistance that can be detected of a wheatstone brigde given that Resistance 1=Resistance 2= varrying Resistance 3=Resistance 4=1000 ohms and a galvanometer can detect as low as 0.1 milliamperes?

Answer 1

As all resistance given are equal connected in Wheatstone bridge

Therefore the condition of balance

$R_1*R_3= R_2*R_4$ is satisfied

No current flows through galvanometer hence the internal resistance of galvanometer is infinite for this case.