The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 \Omega$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Current Electricity

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of

15Ω

resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Answer: (C)

With the galvanometer branch removed, ABC carries

100 + 10 = 110 Ω

and ADC carries

60 + 5 = 65 Ω

across the same 10 V. Taking

V_{A} = 10

V and

V_{C} = 0

V_{B} = 10 - \frac{10}{110} (100) = \frac{10}{11} V, V_{D} = 10 - \frac{10}{65} (60) = \frac{10}{13} V

so the open-circuit voltage across BD is

\frac{10}{11} - \frac{10}{13} = \frac{20}{143} = 0.1399

V.
Replacing the source by a short joins A to C, and the resistance looking back into the bridge from B and D is

R_{th} = \frac{100 \times 10}{110} + \frac{60 \times 5}{65} = \frac{100}{11} + \frac{60}{13} = 13.71 Ω

The galvanometer therefore carries

I_{g} = \frac{0.1399}{13.71 + 15} = 4.87 \times 1 0^{- 3} A

that is, a current of

4.87

mA.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option