Answer to Question #216334 in Electrical Engineering for Fredy

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.