In the given circuit,the galvanometer shows z | vedclass

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

Question

(D) For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal. The equivalent resistance of the parallel combination of

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal. The equivalent resistance of the parallel combination of $100 \ \Omega$ and $R$ is $R_{eq} = \frac{100 	imes R}{100 + R}$.The condition for a balanced bridge is:$\frac{100}{200} = \frac{R_{eq}}{40}$Substituting $R_{eq}$:$\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 = \frac{400}{16} = 25 \ \Omega$.

In the given circuit,the galvanometer shows zero deflection. What is the value of $R$ in $\Omega$?

Similar Questions

The figure shows a network of four resistors and three batteries. Which of the batteries is getting charged?

The resistances in the Wheatstone bridge circuit shown in the diagram are all different,and the current through the galvanometer is zero. If all thermal effects are negligible,the current through the galvanometer will not be zero when:

In the given circuit diagram,find the current passing through the wire $CD$ (in ampere).

In the following network,the current flowing through the $15\Omega$ resistance is: (in $A$)

The potential difference $(V_A - V_B)$ between the points $A$ and $B$ in the given figure is (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module