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In a balanced Wheatstone bridge with resistors $R_1, R_2, R_3, R_4$ (galvanometer across the B-D diagonal), the balance condition for zero galvanometer current is
A$\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$
B$R_1 R_2 = R_3 R_4$
C$R_1 + R_2 = R_3 + R_4$
D$\dfrac{R_1}{R_3} = \dfrac{R_4}{R_2}$
Answer & Solution
Correct answer: A. $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$
1. At balance, no current flows through the galvanometer ($I_g = 0$).
2. Applying the loop rule to the two loops gives $I_1 R_1 = I_2 R_2$ and $I_1 R_3 = I_2 R_4$.
3. Dividing the two relations yields $R_1/R_2 = R_3/R_4$.
4. So A; the product and sum forms (B, C) are not the balance condition.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.20_
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