A meter bridge with two resistances R₁ and R₂ as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 Ω resistance is connected in parallel to R₂. The values of resistances R₁ and R₂ are ______. (a) R₂ = 8 Ω, R₁ = 16/3 Ω (b) R₂ = 12 Ω, R₁ = 12/3 Ω (c) R₂ = 4 Ω, R₁ = 4/3 Ω (d) R₂ = 16 Ω, R₁ = 16/3 Ω

Answer: (a). Initially (R₁)/(R₂) = 40/60 = 2/3. With 16 Ω in parallel, balance at 50 cm means R₁ = (16R₂)/(16+R₂). 2R₂/3 = (16R₂)/(16+R₂) ⇒ 32 + 2R₂ = 48 ⇒ R₂ = 8 Ω. R₁ = 2/3 × 8 = 16/3 Ω. So the answer is (a).

A meter bridge with two resistances R₁ and R₂ as shown in figure was… | JEE Main 2026 (22 Jan, Shift 1)

Q16JEE Main 2026 · 22 JAN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge

A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at 40 cm from the point $P$ . The null point changed to 50 cm from the point $P$ , when $16\ \Omega$ resistance is connected in parallel to $R_2$ . The values of resistances $R_1$ and $R_2$ are ______.

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Answer: (a)

Initially $\dfrac{R_1}{R_2} = \dfrac{40}{60} = \dfrac{2}{3}$ .
With $16\ \Omega$ in parallel, balance at 50 cm means $R_1 = \dfrac{16R_2}{16+R_2}$ .
$\dfrac{2R_2}{3} = \dfrac{16R_2}{16+R_2} \Rightarrow 32 + 2R_2 = 48 \Rightarrow R_2 = 8\ \Omega$ .
$R_1 = \dfrac{2}{3} \times 8 = \dfrac{16}{3}\ \Omega$ .
So the answer is (a).

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Wheatstone Bridge & Metre Bridge question from JEE Main 2026 (22 Jan, Shift 1)

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