Question

give the derivation for wheatstone bridge❣️​​

Answer 1

Answer:

Wheatstone Bridge Derivation

The current enters the galvanometer and divides into two equal magnitude currents as I1 and I2. The following condition exists when the current through a galvanometer is zero,

I1P=I2R (1)

The currents in the bridge, in a balanced condition, is expressed as follows:

I1=I3=EP+Q I2=I4=ER+S

Here, E is the emf of the battery.

By substituting the value of I1 and I2 in equation (1), we get

PEP+Q=RER+S

PP+Q=RR+S

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

PR+PS=RP+RQ

PS=RQ

(2)

R=PQ×S

(3)

Equation (2) shows the balanced condition of the bridge while (3) determines the value of the unknown resistance.

In the figure, R is the unknown resistance, and the S is the standard arm of the bridge and the P and Q are the ratio arm of the bridge

Explanation:

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Answer 2

Answer:

Hello...

Here Your Answer....

According to Kirchhoff's circuital law, the voltage drop across a closed loop is zero. Applying Kirchhoff's law in the loop ABDA, the sum of voltage drops across the individual arms of the loop is zero i.e.

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See In Attachment Also....

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