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 10V is maintained across AC.

Solution

Physics sir · 4 steps

The bridge is not balanced, so take C as zero volts and write Kirchhoff's current law at B and D.

1. Check balance. 100/10 = 10 but 60/5 = 12. The ratios differ, so current does flow through the galvanometer and we must find V_B - V_D.
2. Node equations. Take V_A = 10 V and V_C = 0. At B and at D, current coming in equals current going out (the galvanometer is the 15 Ω link BD).

   (10 - V_B)/100 = V_B/10 + (V_B - V_D)/15 ⇒ 53V_B - 20V_D = 30 ; (10 - V_D)/60 = V_D/5 + (V_D - V_B)/15 ⇒ -4V_B + 17V_D = 10

3. Solve. These give V_D = 650/821 V and V_B = 710/821 V, so V_B - V_D = 60/821 V ≈ 0.0731 V.
4. So. I_g = (V_B - V_D)/15 = 4/821 A ≈ 4.87 × 10⁻³ A = 4.87 mA, flowing from B to D. That is option (b).

The idea to keep. In an unbalanced bridge, ground one corner, write current-in equals current-out at the two galvanometer nodes, and the galvanometer current is their potential difference divided by its resistance.

Answer from the NTA final answer key. Solution by PrepWiser.