Four resistors P, Q, R and S form a Wheatstone bridge. The bridge is balanced when \(P/Q=R/S\). It is balanced with P = 2 Ω, Q = 4 Ω and R = 3 Ω. Now P and Q are interchanged. To balance the bridge again, S must be changed to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.5 Ω
A balanced bridge shows no current in the galvanometer. This needs the four resistances to satisfy the balance condition. After the interchange, \( P=4\ \Omega \) and \( Q=2\ \Omega \), while \( R \) stays 3 Ω.
\( \begin{aligned} S' &= \frac{RQ}{P} \\ &= \frac{3\times2}{4} \\ &= 1.5\ \Omega \end{aligned} \)
Before the interchange, \( S \) was \( 3\times4/2 \), which is 6 Ω. Hence, S has to be made four times smaller.
NCERT (2026-27):
Wheatstone Bridge – “… Hence, we obtain the condition”
\( \frac{R_2}{R_1}=\frac{R_4}{R_3} \)\( \qquad [3.64(a)] \)
“This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.”