The diagram above illustrates the conversion of a galvanometer of resistance 29 to an ammeter. The galvanometer gives a full-scale deflection for a current ...

Answer Details

The circuit shown is a shunt or parallel circuit. We know that the full-scale deflection current of the galvanometer, I_G = 10mA = 10 x 10^-3A. Let R be the resistance of the shunt resistor. Since the galvanometer and shunt resistor are in parallel, the total current passing through the circuit is I_T = I_G + I. Here, I is the current passing through the shunt resistor. We also know that the current I should be such that the ammeter should read a full-scale deflection when the current I_T = 1A. According to the question, the resistance of the galvanometer is 29Ω. We can calculate the value of the current I_T when a potential difference of 29V is applied across the galvanometer: I_T = V / (R_G + R) where V is the applied potential difference, R_G is the resistance of the galvanometer, and R is the resistance of the shunt resistor. Since we know that I_T should be 1A when the ammeter reads full-scale deflection, we can set up the following equation: 1 = 29 / (29 + R) Solving for R gives: R = 29Ω x (1 - 1/1) = 0Ω This is not a practical value for a shunt resistor. It means that the ammeter will read zero current regardless of the actual current passing through the circuit. Therefore, we need to select an option which states that the value of R is very small. The correct option is therefore: 2.0 x 10^-3 Ω.