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 ...

You visited us 62 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

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

\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}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

Similar questions

Q. In the adjoining diagram

R_{1} = 10 Ω, R_{2} = 20 Ω, R_{3} = 40 Ω, R_{4} = 80 Ω

and

V_{A} = 5 V, V_{B} = 10 V, V_{C} = 20 V, V_{D} = 15 V

. The current in the resistance

R_{1}

will be

Q. If a polynomial

f (x) = 4 x^{4} - a x^{3} + b x^{2} - c x + 5

(a, b, c ϵ R)

has four positive real roots

r_{1}, r_{2}, r_{3}, r_{4}

such that

\frac{r_{1}}{2} + \frac{r_{2}}{4} + \frac{r_{3}}{5} + \frac{r_{4}}{8} = 1

Then value of a will be

Q. If a polynomial

f (x) = 4 x^{4} - a x^{3} + b x^{2} - c x + 5

(a, b, c \in R)

has four positive real zeros

r_{1}, r_{2}, r_{3}, r_{4}

such that

\frac{r_{1}}{2} + \frac{r_{2}}{4} + \frac{r_{3}}{5} + \frac{r_{4}}{8} = 1

, then value of

a

Q. The potential difference in volt across the resistance

R_{3}

in the circuit shown In figure, is

(R_{1} = 15 Ω, R_{2} = 15 Ω, R_{3} = 30 Ω, R_{4} = 35 Ω)

Q. In fig. the equivalent resistance of the following combination of resistors

r_{1}, r_{2}, r_{3}

and

r_{4}

r_{1} = r_{2} = r_{3} = r_{4} = 2.0 Ω

, between the points A and B is

Join BYJU'S Learning Program

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

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