Wheatstone Bridge

The Wheatstone Bridge is a basic electrical circuit that measures unknown resistances with very high accuracy. It was invented by Samuel Hunter Christie in 1833 and later popularised by Charles Wheatstone in 1843. It is used as a bridge circuit in laboratories and other practical applications where resistance measurements are needed with great precision.

1.0What is Wheatstone Bridge?

Wheatstone Bridge is a device used in electrical circuits that helps measure an unknown resistance with great accuracy. 

2.0Wheatstone Bridge Circuit

The Wheatstone bridge circuit consists of four resistors connected by a galvanometer and a power source in the form of a rectangle. 

1. Four Resistors: The four resistors are connected in the form of a rectangle. The known resistors are three of the four resistors in total, and the fourth is the unknown one whose value has to be found.
2. Power Source: The opposite sides of the rectangle are connected with a constant voltage source to supply the current.
3. Galvanometer: A galvanometer is connected between the two midpoints of the rectangle. It's a device that measures small quantities of current flowing through it.
4. R1, R2, R3, and Rx: These are the known resistance and unknown resistance connected with R1, R2, and R3 in one of the arms of the bridge.

Deflection of Galvanometer

The deflection of a galvanometer is defined as the angular displacement of the needle or pointer of the galvanometer as a result of the flow of current. This displacement is directly proportional to the current passing through the galvanometer. The deflection of the galvanometer is given by: 

$\theta =kI$

Here:

• θ is the deflection of the galvanometer (in radians or degrees).
• k is the constant of the galvanometer, known as the galvanometer's constant (depending on the galvanometer's design and sensitivity, the value of k is determined experimentally).
• I is the current passing through the galvanometer.

3.0Wheatstone Bridge Working Principle 

This important principle of balance is incorporated in the Wheatstone Bridge. When the Bridge is in balance, the current in the galvanometer will be zero. This is achieved only if the ratio of the two resistances in one branch is equal to the ratio of the other two in another branch. Balance condition:

      $\frac{R1}{R2}=\frac{R3}{Rx}$

In this equation:

• R1, R2, and R3 are the known resistors, and
• Rx is the unknown resistor.

Once the Bridge is balanced, the galvanometer shows zero current or, in other words, no voltage difference between its terminals. 

Zero galvanometer current is described as that condition of a circuit where no current flows in the galvanometer. This happens when the potential difference between the two points of the galvanometer comes out to be zero. 

Thus, we can solve for the value of Rx from the following formula:

$Rx=\frac{R3\times R2}{R1}$

4.0Derivation for Balancing Condition of Wheatstone Bridge

Wheatstone bridge basically works on the condition when there is no current flowing through the galvanometer otherwise the condition would unbalanced, and hence the current will be zero. 

When the current through the galvanometer is zero the Wheatstone bridge will be in the balanced condition. And the potential will be equal across the two terminals namely A and B. 

$V_A=V_B$

$I_1{R}_1=I_2{R}_3$

When there is no current in the circuit I₁ will go through R₁ and R₂, similarly, I₂ will go through R₃ and R_x. So we can write: 

$I_1=\frac{V}{R_1+R_2}$ and $I_2=\frac{V}{R_3+R_x}$

So we get, 

$\frac{V}{R_1+R_2}{R}_1=\frac{V}{R_3+R_x}{R}_3$

$\frac{R_1}{R_1+R_2}=\frac{R_3}{R_3+R_x}$

$R_1(R_3+R_x)=R_3(R_1+R_2)$

$R_1{R}_x=R_3{R}_2$

$\frac{R_1}{R_3}=\frac{R_2}{R_x}$

Process of Balancing the Wheatstone Bridge: 

1. Setup: Connect resistors R₁​, R₂ (known), R₃​ (unknown), and R₄​ (variable). Attach a galvanometer between the midpoints of opposite resistors.
2. Adjust Variable Resistance: Change R4​ and check the galvanometer reading.
3. Achieve Zero Current: The bridge is balanced when the galvanometer shows zero current.
4. Balance Condition: Balance the bridge using condition: 

$\frac{R_1}{R_2}=\frac{R_3}{R_4}$

5. Calculate unknown resistance: after balancing the bridge calculate the unknown resistance. 
6. Final Check: Ensure Zero current in the galvanometer for balance.

5.0Conditions for Balance

The Wheatstone will be able to perform its duty correctly and display the true value of the unknown if only it happens to be in its balanced condition. The potential difference between the two midpoints with the galvanometer joined, therefore, must be zero. In such a case, the current does not pass through the galvanometer. For it to reach equilibrium, these steps are conducted:

• The known resistors, R1, R2, and R3, are varied until the galvanometer has no deflection. At this point, the current flowing between the two branches of the bridge is equal, and the system is in balance.
• The balance condition is fulfilled, and the unknown resistance can be determined from the formula above.

6.0Applications of Wheatstone Bridge

Usage of Wheatstone Bridge is unlimited, and it is adopted in a variety of locations where precise measurement of resistances is required. It is utilized in the following common ways: 

1. Measurement of Resistance: The main application of the Wheatstone is in the accurate measurement of the resistance of an unknown resistor. 
2. Calibration of Instruments: The Wheatstone Bridge is also used to calibrate devices such as ammeters, voltmeters, and ohmmeters. 
3. Temperature Sensing (Thermistors): The Wheatstone can also be used to measure small changes in temperature. A thermistor resistor whose resistance varies with temperature is added as one of the resistors, and the change in resistance can be measured as a change in voltage or current.
4. Wheatstone Bridge Capacitor: Although it has been used traditionally for measuring resistances, the Wheatstone Bridge can easily be modified to measure capacitance in some circuits. 

7.0Solved Examples

Problem 1. In a Wheatstone Bridge, the following resistances are connected:

R1=10$\Omega$,

R2=20$\Omega$

R3=15$\Omega$

The galvanometer shows zero deflection, indicating the bridge is balanced. Find the value of the unknown resistance $R_4$.

Solution: For the Bridge to be balanced, the condition is:

$\frac{R_1}{R_2}=\frac{R_3}{R_4}$

Substitute the given values:

$\frac{10}{20}=\frac{15}{R_4}$

Now solve for R4:

$R_4=\frac{15\times 20}{10}=30\Omega$

Thus, the value of the unknown resistance $R_4$ is $30\Omega$

Problem 2: In a Wheatstone Bridge, the resistances are as follows:

• R1=5 Ω
• R2=10 Ω
• R3=15 Ω
• R4=x Ω (unknown resistance)

When the bridge is unbalanced, the voltage across the galvanometer is measured to be 2.5 V. The battery applied is 10 V. Find the value of x, the unknown resistance.

Solution: For an unbalanced condition in a Wheatstone bridge, the formula to find voltage or potential difference across the galvanometer is given by: 

$V_G=(\frac{R_2}{R_1+R_2}-\frac{R_4}{R_3+R_4})\times V$

$2.5=(\frac{10}{15}-\frac{x}{15+x})\times 10$

$0.25=(\frac{2}{3}-\frac{x}{15+x})$

$\frac{x}{15+x}=\frac{2}{3}-0.25=\frac{2}{3}-\frac{1}{4}=\frac{1}{4}$

$\frac{x}{15+x}=\frac{1}{4}$

$4x=15+x$

$3x=15$

$x=5\Omega$

Problem 3: In a Wheatstone Bridge, the following resistances are given:

• R1=150$\Omega$
• R2=250$\Omega$
• R3=75$\Omega$
• R4= 100$\Omega$

The applied voltage is 20 V, and the galvanometer is connected between the midpoints of the two branches. The deflection on the galvanometer is recorded as 5 divisions. Calculate the deflection on the galvanometer when the bridge is balanced.

Solution: First, let’s check if the bridge is balanced or not: 

$\frac{R_1}{R_2}=\frac{R_3}{R_4}$

$\frac{150}{250}=\frac{75}{100}$

$\frac{3}{5}\ne \frac{3}{4}$

Hence, the bridge is not balanced so to find the voltage of the galvanometer is: 

$V_G=(\frac{R_2}{R_1+R_2}-\frac{R_4}{R_3+R_4})\times V$

$V_G=(\frac{250}{150+250}-\frac{100}{75+100})\times 20$

$V_G=(\frac{250}{400}-\frac{100}{175})\times 20$

$V_G=(\frac{5}{8}-\frac{4}{7})\times 20$

$V_G=\frac{35-32}{56}\times 20$

$V_G=\frac{15}{14}\approx 1.07V$

The deflection of any galvanometer is directly proportional to the potential difference across it. Hence, 

Deflection = $1.07\times 5=5.35$ divisions