A Wheatstone bridge is a configuration of resistors and a sensitive current
meter, called a Galvanometer, that is used to determine the resistance of
an unknown resistor. In the Wheatstone bridge shown here, find the
value of Rx such that the current through Galvanometer G is zero.
How does this configuration work?
Thanks.
The two concepts that are essential here are Kirchoff's junction rule and the fact that parallel pathways experience the same voltage drop.
If the galvanometer reads 0V, then the voltage drop across the 10-ohm resistor is equal to the voltage drop across the 20-ohm resistor. It also means that the voltage drop across the unknown resistor is equal to the voltage drop across the 5-ohm resistor. This gives us:
I10(10 ohms) =
I20(20 ohms)
and
Iunknown(X ohms) =
I5(5 ohms)
We also know that because the 10-ohm resistor is in series with the unknown resistor and the galvanometer reads 0V, that currents through those two resistors are equal. This gives us:
I10 =
Iunknown
and
I20 =
I5
This pool of information allows us to solve for the resistance of the unknown resistor.
Start with the first half where:
I20/
I10 = R
10/R
20 = 10/20 = 1/2 = 0.5
We know from the series relationship that
I20/
I10 =
I5/
Iunknown = 0.5
Plugging into
I5/
Iunknown = R
unknown/R
5 we get:
R
unknown/R
5 = 0.5.
From this we get R
unknown = 0.5R
5 = 0.5(5) = 2.5.
That's the math of it.
In a simple perspective, we basically know that the ratio of the first resistors (20 and 10) must equal the ratio of the second resistors (5 and unknown) as long as the galvanometer reads 0, meaning that the voltage drops are equal for the various segments.
In a typical wheatstone bridge, the 5-ohm resistor actually represents a variable resistor that is adjucted until the galvanometer reads 0.
If you have access to the BR in-class passages, there a really good example of this device and an experiment using it.
