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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Obtain the condition for bridge balance in Wheatstone’s bridge.

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प्रश्न

Obtain the condition for bridge balance in Wheatstone’s bridge.

Obtain the balancing conditions in the case of Wheatstone’s bridge.

व्युत्पत्ती
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उत्तर १

An important application of Kirchhoff’s rules is the Wheatstone Bridge. It is used to compare resistances and also helps in determining the unknown resistance in an electrical network. The bridge consists of four resistances, P, Q, R, and S, connected. A galvanometer G is connected between the points B and D. The battery is connected between points A and C. The current through the galvanometer is IG, and its resistance is G.


      Wheatstone’s bridge

Applying Kirchhoff’s current rule to junction B,

I1 – IG – I3 = 0    ...(1)

Applying Kirchhoff’s current rule to junction D,

I2 + IG – I4 = 0    ...(2)

Applying Kirchhoff’s voltage rule to loop ABDA,

I1P + IGG – I2R = 0    ...(3)

Applying Kirchhoff’s voltage rule to loop ABCDA,

I1P + I3Q – I4S – I2R = 0    ...(4)

When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through the galvanometer (IG = 0).

Substituting IG = 0 in equations (1), (2), and (3), we get

I1 = I3    ...(5)

I2 = I4    ...(6)

I1P = I2R    ...(7)

Substituting equations (5) and (6) in equation (4),

I1P + I1Q – I2R = 0

I1(P + Q) = I2 (R + S)    ...(8)

Dividing equation (8) by equation (7), we get

`(P + Q)/P = (R + S)/R`

`11 + Q/P = 1 + S/R`

⇒ `Q/P = S/R`

`P/Q = R/S`    ...(9)

This is the bridge balance condition. Only under this condition, the galvanometer shows null deflection. Suppose we know the values of two adjacent resistances; the other two resistances can be compared. If three of the resistances are known, the value of the unknown resistance (the fourth one) can be determined.

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उत्तर २

In a Wheatstone bridge at balance, no current flows through the galvanometer. Hence, the potential at junction points is equal.

Let resistances be P, Q, R, and S.

By using the given condition:

`P/Q = R/S`

⇒ PS = QR

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Notes

Students can refer to the provided solutions based on their preferred marks.

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पाठ 2: Current Electricity - Evaluation [पृष्ठ १२०]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 2 Current Electricity
Evaluation | Q III. 6. | पृष्ठ १२०

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