Wheatstone Bridge

The Wheatstone Bridge is a circuit arrangement used to determine an unknown resistance with high accuracy by comparing it with known resistances. 

It is based on the condition of null deflection, in which no current flows through the galvanometer.

An arrangement of four resistors used to measure the resistance of one of them in terms of the other three, invented by Samuel Hunter Christie in 1833 and made famous by Sir Charles Wheatstone, is called a Wheatstone bridge.

The condition of the Wheatstone bridge under which the galvanometer shows zero (null) deflection, i.e., I_g = 0, is called the balance condition of the bridge.

Balance condition (when I_g = 0):

In the standard Wheatstone bridge arrangement: 

• Four resistors form the four arms of the bridge. 
• One diagonal acts as the battery arm. 
• The other diagonal acts as the galvanometer arm. 
• One of the resistors may be unknown and is calculated using the balance condition. 

Standard Arm Labels

The bridge works on the principle that when the potentials at the two galvanometer junctions become equal, no current flows through the galvanometer. 

At this stage, the bridge is balanced, and the ratio of resistances in one pair of arms equals the ratio in the other pair of arms. 

Stepwise Logic

1. Current from the battery divides into different branches of the bridge. 
2. The resistances are adjusted or chosen suitably. 
3. When the galvanometer shows zero deflection, the two junctions connected to it are at the same potential. 
4. The balance condition is then applied to calculate the unknown resistance.

Find the current through the galvanometer Ig.

Given

• AB = 100 Ω, BC = 10 Ω, CD = 5 Ω, DA = 60 Ω
• Galvanometer G = 15 Ω (across BD)
• Battery across AC = 10 V

Steps

1. Assume: currents in main branches = I_1​ (AB – BC) and I₂​ (AD – DC), galvanometer current = I_g.
2. Write KVL in loop ABD: get equation (3.65a) in I₁, I₂, I_g.
3. Write KVL in loop BCD: get equation (3.65b).
4. Write KVL in the outer loop with the battery: get equation (3.65c).
5. Solve the three equations (eliminate I₁, I₂​) to get Ig ≈ 4.87 mA.

Key idea: Unbalanced bridge → must use full KVL equations, not the balance formula.

Find variable resistor Q and output voltage.

Given

• Bridge arms: P, Q, R, S (with given values)
• Q is variable, supply = 30 V
• The bridge is balanced

Steps

1. Use the balance condition:
   \[\frac {P}{Q}\] = \[\frac {R}{S}\] ⇒ Q = \[\frac {PS}{R}\]
   Substitute values → Q ≈ 19.95 kΩ.
2. Find total resistance in Q‑branch: R_branch = Q + other resistor.
3. Current in that branch: I₁ = 30 / R_branch.
4. Potential across Q: V_AD = I₁ × Q ≈ 24.58 V.
5. Do the same for the other branch: find current I₂, then V_AB ≈ 24.58 V.
6. Output voltage: Vout = V_AB − V_AD = 0 V (because the bridge is balanced).

Key idea: Balanced bridge → use ratio formula to get Q, and potentials at the two midpoints become equal, so output = 0.

The source material links the Wheatstone bridge to several practical and derived devices. 

Main Applications

• Measurement of an unknown resistance. 
• Working principle of the meter bridge. 
• Use in post office box arrangements. 
• Use in Carey-Foster’s bridge. 
• Calibration-related applications for electrical measuring instruments such as voltmeters and ammeters. 

A balanced Wheatstone bridge can be compared to a physical balance with equal moments on both sides. When both sides are perfectly matched, the indicator shows no tilt, just as the galvanometer shows no deflection.