JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Current Electricity

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

16Ω

resistance is connected in parallel to

R_{2}

. The values of resistances

R_{1}

and

R_{2}

are ______.

Answer: (C)

Solution

At balance the metre bridge gives

\frac{R _{1}}{R _{2}} = \frac{40}{60} = \frac{2}{3}

. Connecting

16 Ω

in parallel with

R_{2}

reduces that arm to

\frac{16 R _{2}}{R _{2} + 16}

, and the null now sits at the mid-point, so

R_{1} = \frac{16 R _{2}}{R _{2} + 16}

. Substituting

R_{1} = \frac{2}{3} R_{2}

gives

\frac{2}{3} (R_{2} + 16) = 16

, hence

R_{2} = 8 Ω

and

R_{1} = \frac{16}{3} Ω.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option