The given Wheatstone bridge shows no deflecti | vedclass

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

Question

(A) For a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal,i.e.,$\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}}$.Here,$R_{

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) For a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal,i.e.,$\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}}$.Here,$R_{AB} = 100 \ \Omega$ and $R_{AD} = 200 \ \Omega$.The resistance between $B$ and $C$ is the parallel combination of $100 \ \Omega$ and $R$,which is $R_{BC} = \frac{100 	imes R}{100 + R}$.The resistance $R_{CD} = 40 \ \Omega$.Substituting these values into the balanced bridge condition:$\frac{100}{200} = \frac{\frac{100R}{100+R}}{40}$$\frac{1}{2} = \frac{100R}{40(100+R)}$$\frac{1}{2} = \frac{10R}{4(100+R)}$$4(100+R) = 20R$$400 + 4R = 20R$$16R = 400$$R = 25 \ \Omega$.

The given Wheatstone bridge shows no deflection in the galvanometer joined between the points $B$ and $D$ (see figure). Calculate the value of $R$ in $\Omega$.

Similar Questions

In the given circuit,find the potential difference between $P$ and $Q$ in $V$.

Which of the following is a wrong statement?

In a Wheatstone bridge,$P = 90\,\Omega$,$Q = 110\,\Omega$,$R = 40\,\Omega$,and $S = 60\,\Omega$. $A$ cell of $4\,V$ emf is connected across the input terminals. The potential difference between the diagonal points $B$ and $C$ (where the galvanometer is connected) is ............. $V$.

The circuit shown is in a steady state. Now,when the switch is closed,the galvanometer shows no deflection. Then the correct relation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module