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

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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.

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