Question

A galvanometer, having a resistance of \(50 \Omega\), gives a full scale deflection for a current of \(0.05 \mathrm{~A}\). The length in meter of a resistance wire of area of cross-section \(2.97 \times 10^{2} \mathrm{~cm}^{2}\) that can be used to convert the galvanometer into an ammeter which can read a maximum of \(5 \mathrm{~A}\) current is (specific resistance of wire $=5 \times 10^{-7} \Omega \mathrm{m}$ ) (A) \(1.5\) (B) 6 (C) 8 (D) 3

Short answer

The length of the resistance wire required to convert the galvanometer into an ammeter that can read a maximum current of 5 A is approximately 6 meters. Therefore, the correct answer is choice (B).

Step-by-step solution

Calculate the required resistance to convert the galvanometer into an ammeter

Given that the galvanometer has a resistance of 50 Ω and a full-scale deflection current of 0.05 A. We need to convert it into an ammeter that can read a maximum current of 5 A. Let's call the required resistance R_shunt which will be connected in parallel with the galvanometer. Using the formula for resistors in parallel: \[ \frac{1}{R_T} = \frac{1}{R_G} + \frac{1}{R_{shunt}} \] where, R_T = total resistance of the parallel combination R_G = resistance of the galvanometer = 50 Ω R_shunt = required resistance to convert the galvanometer To find R_T, we can use the formula for maximum current: I = V / R_T Since the full-scale deflection current in the galvanometer is 0.05 A, when connected in parallel with the required resistor, the total current passing through the combination will be 5 A. Thus, \[ R_T = \frac{V}{I} = \frac{V}{5} \] where V is the voltage across the galvanometer, which can be found using Ohm's law: V = I * R_G = 0.05 * 50 = 2.5 V Substitute the value of V into the equation for R_T: R_T = 2.5 / 5 = 0.5 Ω Now, we can solve for R_shunt using the formula for resistors in parallel: \[ \frac{1}{0.5} = \frac{1}{50} + \frac{1}{R_{shunt}} \] Solving for R_shunt, we get: \[ R_{shunt} = \frac{1}{\frac{1}{0.5} - \frac{1}{50}} = 1.03 \Omega \]

Find the length of the wire using specific resistance, area of cross-section, and resistance

We are given the specific resistance (ρ) of the wire as \(5 \times 10^{-7} \Omega m\) and the area of cross-section (A) as \(2.97 \times 10^{-2} cm^2 = 2.97 \times 10^{-6} m^2\). We can now use these values and the value of R_shunt to find the length (L) of the wire. Using the formula for resistance, we get: \[ R_{shunt} = \rho \cdot \frac{L}{A} \] Rearrange the formula to find the length: \[ L = \frac{R_{shunt} \cdot A}{\rho} \] Substitute the values into the equation: \[ L = \frac{1.03 \times 2.97 \times 10^{-6}}{5 \times 10^{-7}} \] Calculate the length: \[ L \approx 6.04 m \] So, the length of the resistance wire required to convert the galvanometer into an ammeter that can read a maximum current of 5 A is approximately 6 meters. Therefore, the correct answer is choice (B).