The condition such that the potential difference- V A - V B -0-A- R1-D0-R2-nB- R 3 C- RD- R 1- R 3- R 4- R 4

In a simple setup of Wheatstone bridge as shown below, R_1/R=R_2/R_s There is a simple trick so as to not get confused. Just take the product of resistors ac ...

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Standard XII

Physics

Wheatstone Bridge

The condition...

Question

The condition such that the potential difference, $V_{A} - V_{B} = 0$ .

\frac{R_{1}}{R_{4}} = \frac{R_{2}}{R_{3}}

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\frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}}

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\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{R_{1}}{R_{3}} = \frac{R_{4}}{R_{2}}

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Solution

The correct options are
B $\frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}}$
C $\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$
In a simple setup of Wheatstone bridge as shown below, $\frac{R_{1}}{R} = \frac{R_{2}}{R_{s}}$

There is a simple trick so as to not get confused. Just take the product of resistors across the galvanometer, in an 'x' fashion, and equate. For example in the setup as shown, $R_{1} R_{s} = R_{2} R$

So accordingly for this setup, $R_{1} R_{4} = R_{2} R_{3}$
Which when rearranged, gives
$\frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}}$ and also
$\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$

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The condition such that the potential difference, VA−VB=0.

The condition such that the potential difference, $V_{A} - V_{B} = 0$ .