In the gaps of a meter bridge,two unknown res | vedclass

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

Question

(B) In the first case,according to the principle of the meter bridge: $\frac{A}{B} = \frac{3}{4} \implies 4A = 3B \implies A = \frac{3B}{4} \dots (i)$

Vedclass · Accepted Answer

$30 \Omega, 40 \Omega$

Vedclass · Answer

(B) In the first case,according to the principle of the meter bridge: $\frac{A}{B} = \frac{3}{4} \implies 4A = 3B \implies A = \frac{3B}{4} \dots (i)$In the second case,each resistance is increased by $40 \Omega$: $\frac{A+40}{B+40} = \frac{7}{8}$Cross-multiplying gives: $8(A+40) = 7(B+40) \implies 8A + 320 = 7B + 280 \dots (ii)$Substituting equation $(i)$ into equation $(ii)$:$8(\frac{3B}{4}) + 320 = 7B + 280$$6B + 320 = 7B + 280$$B = 320 - 280 = 40 \Omega$Now,substituting $B = 40 \Omega$ into equation $(i)$:$A = \frac{3 	imes 40}{4} = 30 \Omega$Therefore,the values of $A$ and $B$ are $30 \Omega$ and $40 \Omega$ respectively.

Similar Questions

In a meter bridge,the balancing length from the left end is found to be $25 \ cm$. The value of the unknown resistance is (assume,standard resistance of $1 \ \Omega$ is in the right gap). (in $\Omega$)

In a meter bridge,the point $D$ is the neutral point (null point) as shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module