As shown in the figure, the resistance of a galvanometer G can be found by the half-deflection method. Here the resistance R₂ is adjusted such that when the key K is closed the deflection in the galvanometer becomes half of the value as compared to when K is open. Half-deflection is obtained at R₂ = 4 Ω and thus the galvanometer resistance is found to be 6 Ω . In this half-deflection condition the current (in mA) through the resistor R₁ is:

Step-by-step solution

Let the resistance of the galvanometer be G and the series resistance be R₁ . When the key K is open, the current through the galvanometer is: I₁ = V R₁ + G When the key K is closed, the total resistance of the circuit becomes: R_ eq = R₁ + G R₂ G + R₂ The total current from the battery is I = V R_ eq . The current through the galvanometer is: I₂ = I × R₂ G + R₂ = V R₁ + G R₂ G + R₂ × R₂ G + R₂ = V R₂ R₁(G + R₂) + G R₂ In the half-deflection condition, I₂ = I₁ 2 : V R₂ R₁(G + R₂) + G R₂ = V 2(R₁ + G) Cross-multiplying and simplifying: 2 R₂ (R₁ + G) = R₁(G + R₂) + G R₂ 2 R₁ R₂ + 2 G R₂ = R₁ G + R₁ R₂ + G R₂ R₁ R₂ + G R₂ = R₁ G R₁(G - R₂) = G R₂ R₁ = G R₂ G - R₂ Given G = 6 Ω and R₂ = 4 Ω , we can find R₁ : R₁ = 6 × 4 6 - 4 = 24 2 = 12 Ω In the half-deflection condition (key K closed), the total equivalent resistance of the circuit is: R_ eq = 12 + 6 × 4 6 + 4 = 12 + 2.4 = 14.4 Ω The current through the resistor R₁ is the total current from the battery: I = V R_ eq = 10 14.4 = 100 144 = 25 36 A Converting the current to mA: I = 25 36 × 1000 mA = 25000 36 mA ≈ 694.44 mA