Question

prove the whistone bridge rule through ohm's law​

Answer 1

Answer:

(See the diagram in attachment)

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,

$I_{1}P = I_{2} R$     --eq (1)

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

$I_{1 }= I_{3 }=\frac{E}{P+Q} I_{2}= I_{4 }= \frac{E}{R+S}$

Here, E is the emf of the battery.

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

PE/P+Q=RE/R+S

P/P+Q=R/R+S

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

PR+PS=RP+RQ

PS=RQ                             --eq (2)

R=PS/Q                         -- eq (3)

Hence Proved....

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