Wheatstone bridge is a bridge where four resistances are used. First and second resistance are connected in series; third and fourth resistance are also connected in series. Then the first and second two combinations are connected in parallel make Wheatstone bridge.

**Figure:** Wheatstone Bridge Circuit

In figure P, Q, R, S are four resistances. P and Q resistors are connected in series at B point; R and S resistors are connected in series also at D point. Combination of P-Q and R-S are connected in parallel at A-C point. E is supply voltage source. Total current I divides into I_{1}, I_{2} at A point according to Kirchoff’s Current divider rule. The potential of B and D points must be same according to Wheatstone bridge principle. So a galvanometer is placed between B and D terminal to check whether the potential is same or not. When there is any potential difference these two points then the bridge is called unbalanced. If potential are same it is balanced or stable circuit.

As V_{B} = V_{D} I_{1} current flows via Q resistor and I_{2} current flows through S resistor. There is no current flow through galvanometer at ideal wheatstone bridge so I_{g} = 0. Finally I_{3}, I_{4} current meet at C point which sum is total current I.

Principal of Wheatstone bridge is ( P/Q )= ( R/S ).The ratio of first and second resistance value is equal to the ratio of third and fourth resistance value.

At A point, I = I_{1} + I_{2}

At B point, I_{1} = I_{g} + I_{3}

I_{1} = I_{3} [ as I_{g}= 0 ]

At D point, I_{2} = I_{g} + I_{4}

I_{2} = I_{4} [ as I_{g}= 0 ]

Potential of A & B points are V_{A} and V_{B} and potential of C & D points are V_{C} and V_{D}.

There is no potential difference between V_{B} and V_{D}.

Hence, V_{B} = V_{D}.

For P,Q resistors using Ohm’s law V = IR we get,

This is the equation of Wheatstone bridge. Ratio of first & second resistance is equal to third & fourth resistance. When the ratio is equal then the bridge is called balanced if the ratio resistance are not equal then the bridge is called unbalanced.

## How to balance Wheatstone Bridge when unbalanced

Let’s take two examples to understand how to fix unbalanced Wheatstone Bridge.

## Wheatstone bridge example

*Example 1:** A Wheatstone bridge has four value resistance are 8**Ω**, 12**Ω, 16Ω and 20Ω. It is unbalanced. How a resistance should be connected with fourth arm to balance the bridge?*

**Answer:**

**Figure:** Unbalanced Wheatstone bridge Circuit

Here,

First arm resistance, P = 8Ω

Second arm resistance, Q = 12Ω

Third arm resistance, R = 16Ω

Fourth arm resistance, S_{1} = 20Ω

Essential resistance should be connected with fourth arm resistance, S_{2 }= ?

The Wheatstone bridge is unbalanced here because of wrong value of fourth resistance S_{1}. Connecting resistance S_{2} with S_{1} will solve the problem and make the bridge balanced but the question is how should be connected in series or parallel and how much value. Let’s find it.

We know,

If the value of fourth resistance remains 24Ω, the bridge will be balanced.

Expected fourth resistance value is S = 24Ω but here connected value is less than expected value for balanced circuit S_{1} = 20Ω.

So it is clear that fourth resistance value should be increased can balance the bridge. Series connection of a resistance with it twill balance the circuit.

S = S_{1} + S_{2}

or 24 = 20 + S_{2}

S_{2} = 4Ω

**Figure:** balanced Wheatstone bridge Circuit

We can say that 4Ω resistance should be connected in series with fourth arm resistance S_{1} to balance Wheatstone Bridge.

*Example 2:** A Wheatstone bridge has four arms with 5Ω, 15Ω, 20Ω and 100Ω resistance. The bridge is not balanced. What value of resistance should be connected with 4 ^{th} arm and how to connect resistance to balance the bridge?*

**Answer:**

**Figure:** unbalanced Wheatstone bridge circuit

Here,

First arm resistance, P = 5Ω

Second arm resistance, Q = 15Ω

Third arm resistance, R = 20Ω

Fourth arm resistance, S_{1} = 100Ω

Essential resistance should be connected with fourth arm resistance, S_{2 }= ?

We know,

If the 4^{th} arm resistance value S = 60Ω, the bridge will be balanced. The bridge is unbalanced cause of more resistance bearing 4^{th} arm than 60Ω which is S_{1} = 100Ω. To decrease the value of 4^{th} arm another resistance should be connected in parallel to balance the circuit. Let’s find out.

We know from parallel connection,

**Figure:** Balanced Wheatstone bridge in parallel connection

If we connect 150Ω resistance with 4^{th} arm in parallel then the Wheatstone bridge will be balanced.