In a Wheatstone bridge, the four arms have resistances AB = 2 Ω, BC = 4 Ω, AD = 6 Ω and DC = 12 Ω. A 12 V battery of negligible internal resistance is connected between A and C. A galvanometer is connected between B and D. The potential difference across the arm AD is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4 V
Here \( AB/BC=2/4 \) and \( AD/DC=6/12 \). Both ratios are equal. So the bridge is balanced, and no current flows through the galvanometer. The battery current then divides between the two paths ABC and ADC. The full 12 V acts across the path ADC of resistance 18 Ω.
\( \begin{aligned} I_{ADC} &= \frac{12}{6+12} \\ &= \frac{2}{3}\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} V_{AD} &= I_{ADC}\times6 \\ &= \frac{2}{3}\times6 \\ &= 4\ \mathrm{V} \end{aligned} \)
The drop across AB is also 4 V. So B and D are at the same potential, as a balanced bridge needs.
NCERT (2026-27):
Wheatstone Bridge – “… Hence, we obtain the condition”
\( \displaystyle \frac{R_2}{R_1}=\frac{R_4}{R_3} \)\( \qquad [3.64(a)] \)
“This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.”