Answer to Question #214001 in Electrical Engineering for Fredy

Question

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?

Answer 1

Calculating the internal resistance across the galvanometer

R1 and R2 are in parallel, so their effective resistance is

$R_{12}=\frac{R_1\cdot R_2}{R_1+R_2}\\ R_{12}=\frac{1000\cdot 1000}{1000+1000}\\ R_{12}=500\Omega$

R3 and R4 are also in parallel. Their effective resistance is

$R_{34}=\frac{R_3\cdot R_4}{R_3+R_4}\\ R_{34}=\frac{1000\cdot 1000}{1000+1000}\\ R_{34}=500\Omega$

Internal resistance is 1000ohms

R12 is now in series with R13 to give the value of the internal resistance

$R_{int}=R_12+R_34=500+500=1000\Omega$