To solve the problem, we need to understand how a galvanometer can be converted into an ammeter using a shunt resistor. Let's break down the steps to find the current through the galvanometer. ### Step 1: Understanding the Circuit When a galvanometer is converted into an ammeter, a shunt resistor (in this case, 0.05Ω) is connected in parallel with the galvanometer. The purpose of the shunt is to allow most of the current to bypass the galvanometer, while a small portion of the current flows through it. ### Step 2: Given Values - Resistance of the galvanometer, \( R_g = 5 \, \Omega \) - Resistance of the shunt wire, \( R_s = 0.05 \, \Omega \) - Total current that the ammeter can read, \( I = 2 \, \text{A} \) ### Step 3: Current Division in Parallel Resistors In a parallel circuit, the voltage across both resistors (the galvanometer and the shunt) is the same. We can use the current division rule to find the current through the galvanometer (\( I_g \)). The total current \( I \) is divided between the galvanometer and the shunt: \[ I = I_g + I_s \] where \( I_s \) is the current through the shunt. ### Step 4: Using Ohm's Law The voltage across the galvanometer (\( V_g \)) can be expressed using Ohm's Law: \[ V_g = I_g \cdot R_g \] The voltage across the shunt (\( V_s \)) is: \[ V_s = I_s \cdot R_s \] Since \( V_g = V_s \), we can set these two equations equal to each other: \[ I_g \cdot R_g = I_s \cdot R_s \] ### Step 5: Expressing \( I_s \) From the total current equation, we can express \( I_s \) in terms of \( I \) and \( I_g \): \[ I_s = I - I_g \] ### Step 6: Substituting \( I_s \) into the Voltage Equation Substituting \( I_s \) into the voltage equation gives: \[ I_g \cdot R_g = (I - I_g) \cdot R_s \] Now, substituting the known values: \[ I_g \cdot 5 = (2 - I_g) \cdot 0.05 \] ### Step 7: Solving for \( I_g \) Expanding the equation: \[ 5 I_g = 0.1 - 0.05 I_g \] Now, combine like terms: \[ 5 I_g + 0.05 I_g = 0.1 \] \[ 5.05 I_g = 0.1 \] Now, divide both sides by 5.05: \[ I_g = \frac{0.1}{5.05} \approx 0.0198 \, \text{A} \approx 19.8 \, \text{mA} \] ### Step 8: Rounding to the Nearest Option Rounding 19.8 mA gives us approximately 20 mA. Therefore, the current through the galvanometer is: **Correct Option: C. 20 mA** ### Step 9: Why Other Options Are Incorrect - **Option A (2 mA)**: This is too low. The current through the galvanometer is significantly higher than this value. - **Option B (10 mA)**: This is also too low. The calculated current is much higher than 10 mA. - **Option D (25 mA)**: This is too high. The calculated current is less than this value. ### Summary - A galvanometer can be converted into an ammeter using a shunt resistor. - The current through the galvanometer can be calculated using the current division rule and Ohm's Law. - The final calculated current through the galvanometer is approximately 20 mA. ### Revision Summary - Understand the role of the shunt resistor in a galvanometer-to-ammeter conversion. - Use Ohm's Law and the current division rule to find the current through the galvanometer. - Be careful with unit conversions and rounding when interpreting results. - Always check the reasonableness of your answer against the given options.