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In a Wheatstone bridge the unknown resistance is placed in the fourth arm. Known resistances $R_1$ and $R_2$ are kept in the first and second arms while $R_3$ is varied until the galvanometer shows null deflection. The unknown resistance $R_4$ is then
A$R_4 = R_3 \dfrac{R_1}{R_2}$
B$R_4 = R_3 \dfrac{R_2}{R_1}$
C$R_4 = \dfrac{R_1 R_2}{R_3}$
D$R_4 = R_1 + R_2 - R_3$
Answer & Solution
Correct answer: B. $R_4 = R_3 \dfrac{R_2}{R_1}$
1. At balance the bridge satisfies $R_2/R_1 = R_4/R_3$.
2. Solve for the unknown $R_4$: $R_4 = R_3 (R_2/R_1)$.
3. So B is correct.
4. Option A inverts the known ratio; C and D do not come from the balance condition.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.20_
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