In a Wheatstone bridge, \(R_1\) and \(R_2\) are fixed and \(R_4\) is an unknown resistance. The variable resistance \(R_3\) is changed until the galvanometer shows no deflection. The balance condition is \(R_2/R_1=R_4/R_3\). Balance is first found at \(R_3=10\ \Omega\). When a \(12\ \Omega\) resistor is joined in series with \(R_4\), balance is found at \(R_3=16\ \Omega\). The value of \(R_4\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 20 Ω
With \( R_1 \) and \( R_2 \) fixed, the ratio \( k=R_2/R_1 \) does not change. At balance the resistance in the fourth arm is always \( k \) times \( R_3 \). So both balance readings can be written with the same \( k \).
\( \begin{aligned} R_4 &= 10k \end{aligned} \)
\( \begin{aligned} R_4+12 &= 16k \end{aligned} \)
Subtracting the first relation from the second,
\( \begin{aligned} 12 &= 6k \\ k &= 2 \end{aligned} \)
\( \begin{aligned} R_4 &= 10\times2 \\ &= 20\ \Omega \end{aligned} \)
NCERT (2026-27):
Wheatstone Bridge – “… Keeping known resistances R1 and R2 in the first and second arm of the bridge, we go on varying R3 till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance R4 is given by,”
\( R_4=R_3\frac{R_2}{R_1} \)\( \qquad [3.64(b)] \)