A Wheatstone's bridge is balanced with a resi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The balancing condition for a Wheatstone's bridge is $\frac{P}{Q} = \frac{R_3}{R_4}$,where $R_3$ is the resistance in the third arm and $R_4$ is t

Vedclass · Accepted Answer

$650$

Vedclass · Answer

(B) The balancing condition for a Wheatstone's bridge is $\frac{P}{Q} = \frac{R_3}{R_4}$,where $R_3$ is the resistance in the third arm and $R_4$ is the resistance in the fourth arm $(S)$.Given $R_3 = 625\, \Omega$ and $R_4 = S$,we have:$\frac{P}{Q} = \frac{625}{S}$ ..................... $(1)$When $P$ and $Q$ are interchanged,the new resistance in the third arm becomes $625 + 51 = 676\, \Omega$. The bridge is balanced again,so:$\frac{Q}{P} = \frac{676}{S}$$\frac{P}{Q} = \frac{S}{676}$ ..................... $(2)$Equating $(1)$ and $(2)$:$\frac{625}{S} = \frac{S}{676}$$S^2 = 625 	imes 676$$S = \sqrt{625} 	imes \sqrt{676} = 25 	imes 26 = 650\, \Omega$.

Similar Questions

In a Wheatstone's bridge,three resistances $P, Q$ and $R$ are connected in the three arms,and the fourth arm is formed by two resistances $S_1$ and $S_2$ connected in parallel. The condition for the bridge to be balanced will be

If the Wheatstone bridge with four resistors $R_1, R_2, R_3$ and $R_4$ is balanced,then the correct expression is

The equivalent resistance between $A$ and $B$ in the given circuit is . . . . . . . . (in $\Omega$)

Consider the circuit shown in the figure. The current ${I_3}$ is equal to

In the following Wheatstone bridge,$P/Q = R/S$. If key $K$ is closed,then the galvanometer will show deflection:

Vedclass Test Series

Exam Paper Generator

Online Exam Module