A galvanometer of resistance `50Omega` is converted into an ammeter by connecting a low resistance (shunt) of value `1Omega` in parallel to the galvanometer, S. If full - scale deflection current of the galvanometer is 10 mA, then the maximum current that can be measured by the ammeter is -

Options

A. 500 mA
B. 510 mA
C. 610 mA
D. 490 mA

Answer

To solve the problem, we need to find the maximum current that can be measured by the ammeter when a galvanometer is converted into an ammeter by connecting a shunt resistor in parallel.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer (R_g) = 50 Ω
- Shunt resistance (R_s) = 1 Ω
- Full-scale deflection current of the galvanometer (I_g) = 10 mA = 0.01 A

2. **Determine the Voltage Across the Galvanometer:**
The voltage drop across the galvanometer when it is at full-scale deflection can be calculated using Ohm's Law:
\[ V_g = I_g \times R_g = 0.01 \, \text{A} \times 50 \, \Omega = 0.5 \, \text{V} \]

3. **Set Up the Equation for the Shunt Resistor:**
Since the galvanometer and shunt resistor are in parallel, the voltage across the shunt resistor (V_s) is the same as the voltage across the galvanometer:
\[ V_s = I_s \times R_s \]
where \( I_s \) is the current through the shunt resistor.

4. **Equate the Voltages:**
Set the voltage across the galvanometer equal to the voltage across the shunt:
\[ 0.5 \, \text{V} = I_s \times 1 \, \Omega \]
From this, we can solve for \( I_s \):
\[ I_s = 0.5 \, \text{A} = 500 \, \text{mA} \]

5. **Calculate the Total Current Measured by the Ammeter:**
The total current (I) that the ammeter can measure is the sum of the current through the galvanometer and the current through the shunt:
\[ I = I_g + I_s = 10 \, \text{mA} + 500 \, \text{mA} = 510 \, \text{mA} \]

### Final Answer:
The maximum current that can be measured by the ammeter is **510 mA**.

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