[1] Wheatstone bridge/tesistance bridge/Bridge circuit is a series-parallel circuit, with 4 resistors connected to measure a unknown/null center electrical resistance by balancing 2 legs of a bridge circuit, one leg that includes the unknown component.
The upside of using these is its ability to measure very accurately, compared to circuits like voltage dividers.
It also works similarly as a potentiometer.
When the bridge is balanced, voltage across center point is, for accurate calculation of the unknown resistance based on the values of its other known resistors.
No current flows through it if its resistances ratio is equal.
Usually, the bridge is unbalanced where current flows through the galvanometer and is balanced if no current flows through it, which is done by alterong the known resistance and variable resistance.
[1] In this circuit,
RX is the fixed, yet unknown, resistance being measured.
R1 and R3 are resitors of known resistance.
And R2 is adjusted til the bridge is "balanced without current flow across the galanometer Vg, located at the center between points D and B.
[1] At this point, the potential difference between the 2 midpoints, B and D, are 0. So the ratio between the 2 resistance in the known leg (R2/R1) is the ratio fo teh 2 resistance in the unknown leg (RX/R3).
A the balance point:
R2/R1 = RX/R3 → RX = R3 * R2/R1
[1] Detecting no current via a galvanometer can be done at very precisely. So if R1, R2, and R1 very precisely, then so can RX. Very small changes in RX disrupt the balance and are readily detected.
[1] Alternatively, if R1, R2, and R3 are known, but R2 is unadjustable, the voltage difference across or current flow through the meter can be used to calculate the RX's value, via Kirchhoff's circuit laws. This setup is often used in strain gauge and resistance thermometer measurements, as it's often faster to read voltage levels off a meter than to adjust a resistance to zero the voltage.
[6]
This circuit has the same ratio between resistor 5 and 10, and 1 and 2: 5:10 and 1:2, which is a balanced network.
Since it's balanced, no current goes into 6 Ω resistor, whose current is 0, as if the resistor isn't there. Without it 6 Ω resistor, other resistors are now in paralel and can be solved.
This is unbalanced since R1:R2 = 3:1 and R4:R5 = 6:5.
Note: Arrows shows where the points are between the 2 different forms of the same circuit.
The left circuit isn't a Wheatstone circuit since its middle resistor (1 Ω) is connected to the battery path (vertical in this case) is (pointed by red arrows).
Note: Arrows points where the same points are between the 2 forms of the same circuit.
If 1 Ω was at where the battery path isn't at (horizontal in this case), then it's a Wheatstone circuit, 1 Ω has no current, and other resistors are in parallel.
If 5 Ω has x volts, so does 1 Ω.
If 10 Ω has 2x volts, so does 6 Ω.
9 must be divided in the ratio 1:2.
As 9 divides into 1:2, one part gets 1/3 and another gets 2/3.
5 and 1 Ω resistor have 3 V (has 1/3)
10 and 2 Ω resistors have 6 V (has 2/3).
Opposite points also have the same voltages as shown, therefore no current flows in a balanced Wheatstone circuit's middle circuit (6 Ω in this case).
Find the "? resistor" resistance.
We can replace the center resistor with a galvanometer (G), which detects current.
We can take any of the resistors (2 Ω resistor in this case) as a variable resistor.
Change the variable resistor til the galvanometer hits 0 where the resistor changes to 7 V, in this case.
If the current in the middle is 0, the circuit is balanced,
so, if ratio here is 1:7, then "? resistance" is 5x7 = 35 Ω =.
Wheatstone circuits was developed by Charles Wheatstone
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[3]
[4] The Different Types of Electrical Circuits and Their Uses - Titan
[6] Wheatstone bridge & its logic | Electric current | Physics | Khan Academy - YT
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