Current through the galvanometer in the circuit given below is

The correct answer is Option (1) → 121 mA

The 18 V cell is across A and C, and the galvanometer with the \( 5\,\Omega \) resistor is the branch BD (cell and galvanometer taken as ideal). The quickest route is Thevenin's theorem across B and D.

Remove branch BD and take C at zero potential, so A is at 18 V. Branches ABC (\( 100\,\Omega \) and \( 20\,\Omega \)) and ADC (\( 40\,\Omega \) and \( 30\,\Omega \)) are potential dividers:

    \( \displaystyle V_B = 18\times\frac{20}{100 + 20} = 3\,\mathrm{V} \)

    \( \displaystyle V_D = 18\times\frac{30}{40 + 30} = 7.71\,\mathrm{V} \)

    \( \displaystyle V_{\mathrm{th}} = V_D - V_B = 7.71 - 3 = 4.71\,\mathrm{V} \)

With the cell shorted, the \( 100\,\Omega \) and \( 20\,\Omega \) in parallel are in series with the \( 40\,\Omega \) and \( 30\,\Omega \) in parallel:

    \( \displaystyle R_{\mathrm{th}} = \frac{100\times20}{120} + \frac{40\times30}{70} = 16.67 + 17.14 = 33.81\,\Omega \)

Reconnecting branch BD (\( 5\,\Omega \)):

    \( \displaystyle i_g = \frac{V_{\mathrm{th}}}{R_{\mathrm{th}} + 5} = \frac{4.71}{38.81} = 0.121\,\mathrm{A} = 121\,\mathrm{mA} \)

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27):
Wheatstone Bridge – “… Example 3.7 The four arms of a Wheatstone bridge (Fig. 3.19) have the following resistances: AB = 100 Ω, BC = 10 Ω, CD = 5 Ω, and DA = 60 Ω. A galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.”