How do you simplify Wheatstone Bridge?

Current generally enters the given galvanometer and further divides into two equal magnitude currents which are, I₁ and I₂. This condition is only seen when the current passing via a galvanometer is zero.

Hence, I₁P = I₂R … [1]

In a balanced condition, the currents in the bridge can be represented as:

Where,

• E = emf of the battery.

Now, after replacing the value of I₁ and I₂ in equation [1], we can obtain:

⇒ PE/(P+Q) = RE/(R+S)

⇒ P/(P+Q) = R/(R+S)

⇒ P/(R+S)=R/(P+Q)

⇒ PR+PS = RP+RQ

Thus, PS = RQ … [2]

And, R = PQ × S … [3]

\(\therefore\) Equation [2] demonstrates the balanced condition of the bridge. And, Equation [3] specifies the value of the unknown resistance.

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Related Questions

1. Why should we get the null point in the middle of the Metre bridge wire?
2. How does a bridge circuit work?
3. What Does A Wheatstone Bridge In Balanced Condition Provides A Practical Method For?
4. The Image Shows An Unbalanced Wheatstone Bridge. Find Direction Of Conventional Current Between B And D.
5. Can you find very high resistance accurately with the help of a Metre bridge?
6. Why Carey Foster Bridge Is So Sensitive?
7. Meter Bridge or Slide Wire Bridge is a practical form of?
8. Meter bridge works on the principle of?
9. Why is Wheatstone's bridge more accurate?
10. What Is Null Voltage?
11. How do you solve an unbalanced bridge?

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