{"id":2273,"date":"2022-11-01T10:58:02","date_gmt":"2022-11-01T05:28:02","guid":{"rendered":"https:\/\/www.tutoroot.com\/blog\/?p=2273"},"modified":"2025-01-10T12:19:12","modified_gmt":"2025-01-10T06:49:12","slug":"how-wheatstone-bridge-will-work","status":"publish","type":"post","link":"https:\/\/www.tutoroot.com\/blog\/how-wheatstone-bridge-will-work\/","title":{"rendered":"How Wheatstone Bridge will Work?"},"content":{"rendered":"

To find an unknown resistance in a circuit, the Wheatstone bridge is actively used. And it is very important to solve various types of problems and questions, that involve resistors, and other circuits. So, the students need to understand the workings of Wheatstone Bridge. Therefore, to help you out, here in the article below, we will provide a comprehensive description of Wheatstone Bridge. <\/span>\u00a0<\/span><\/p>\n

Principle of Wheatstone Bridge<\/span><\/b><\/h2>\n

The Basic Principle of Wheatstone Bridge is the Principle of Null Deflection. According to this principle, there is no current present in this circuit, and the ratio of resistances is said to be the same. Moreover, at first, the bridge in the circuit, as shown in the diagram, exists in an unbalanced condition, and they can find the current is indeed flowing through the galvanometer. However, to keep the bridge in a balanced condition, there should not be any current flow in the circuit. However, to achieve this balanced position, it is needed to adjust the variable resistance and known resistance in the circuit.<\/span>\u00a0<\/span><\/p>\n

Construction of Wheatstone Bridge<\/span><\/b><\/h2>\n

\"Construction<\/span><\/p>\n

From the above diagram, as you can see, the Wheatstone Bridge consists of four arms, two known and two unknown resistances, and more importantly a variable resistance. The G symbol located in the center of the circuit represents Galvanometer. The emf source is attached to points a and c, while the galvanometer is connected between points b and d.<\/span>\u00a0<\/span><\/p>\n

Wheatstone Bridge Formula Derivation<\/span><\/b><\/h3>\n

The currents I1 and i2 are divided into two equals in magnitude, because of the galvanometer. If the current in the galvanometer is considered zero, then,\u00a0<\/span>\u00a0<\/span><\/p>\n

\\(I_{1}P= I_{2}R\\)<\/strong><\/p>\n

In balanced condition, the terms and quantities in the balanced conditions are represented as,\u00a0<\/span><\/span>\u00a0<\/span><\/p>\n

\\(I_{1}=I_{3}= \\frac{E}{P+Q} \\)<\/strong><\/p>\n

\\(I_{2}=I_{4}= \\frac{E}{R+S}\\)<\/strong><\/p>\n

Where, E = emf of the battery, and<\/span><\/span><\/p>\n

\\(\\frac{PE}{P+Q} =\\frac{RE}{R+S}\\)<\/strong><\/p>\n

\\(\\frac{E}{P+Q}=\\frac{E}{R+S}\\)<\/strong><\/p>\n

\\(P(R+S)=R(P+Q)\\)<\/strong><\/p>\n

\\(PR+PS=PR+QR\\)<\/strong><\/p>\n

\\(PS=RQ\\)<\/strong><\/p>\n

\\(R= \\frac{P}{Q}S\\)<\/strong><\/p>\n

In this equation,\u00a0<\/span><\/p>\n

R<\/span><\/b> = Unknown Resistance<\/span>\u00a0<\/span><\/p>\n

S<\/span><\/b> = Standard Arm of the Bridge<\/span>\u00a0<\/span><\/p>\n

P\/Q<\/span><\/b> = Ratio Arm of Bridge<\/span>\u00a0<\/span><\/p>\n

Applications of Wheatstone Bridge<\/span><\/b><\/h2>\n