The Wheatstone bridge is an arrangement of four resistances, say \( R_1 \), \( R_2 \), \( R_3 \) and \( R_4 \). The null point condition is given by :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( R_1/R_2 = R_3/R_4 \)
At the null point no current flows through the galvanometer, so its two ends are at the same potential.
The current \( I_1 \) through \( R_1 \) then also flows through \( R_2 \), and the current \( I_2 \) through \( R_3 \) also flows through \( R_4 \). Equal potential drops across the matching arms give
\( I_1 R_1 = I_2 R_3 \)
\( I_1 R_2 = I_2 R_4 \)
Dividing the first equation by the second,
\( \displaystyle \frac{R_1}{R_2} = \frac{R_3}{R_4} \)
This is the balance condition of the Wheatstone bridge. Sums, differences or products of the resistances do not express equal potentials at the galvanometer ends.
NCERT (2026-27):
Summary – “… The Wheatstone bridge is an arrangement of four resistances – R1, R2, R3, R4 as shown in the text. The null-point condition is given by
\( \displaystyle \frac{R_1}{R_2} = \frac{R_3}{R_4} \)
using which the value of one resistance can be determined, knowing the other three resistances.”