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 6 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, $V_{A} - V_{B} = 0$ .