A galvanometer of resistance `40 Omega` gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of `2Omega` is connected is

Options

A. 210mA
B. 155 Ma
C. 420mA
D. 75 Ma

Explanation

To solve the problem step by step, let's break it down:

### Given:
- Resistance of the galvanometer, \( R_g = 40 \, \Omega \)
- Deflection per milliampere = 5 divisions
- Total divisions on the scale = 50 divisions
- Shunt resistance, \( R_s = 2 \, \Omega \)

### Step 1: Calculate the maximum current through the galvanometer
The maximum current that can be measured by the galvanometer can be calculated using the total divisions and the deflection per milliampere.

\[ \text{Maximum current through galvanometer, } I_g = \frac{\text{Total divisions}}{\text{Deflection per mA}} = \frac{50 \, \text{divisions}}{5 \, \text{divisions/mA}} = 10 \, \text{mA} \]

### Step 2: Calculate the total current when a shunt is connected
When a shunt resistance is connected, the total current \( I \) that can pass through the circuit is the sum of the current through the galvanometer \( I_g \) and the current through the shunt \( I_s \).

Using the formula for the current through the galvanometer and shunt:
\[ I = I_g + I_s \]

Where:
\[ I_s = \frac{R_g}{R_s} \cdot I_g \]

### Step 3: Substitute the values
Substituting the values we have:
\[ I_s = \frac{40 \, \Omega}{2 \, \Omega} \cdot 10 \, \text{mA} = 20 \cdot 10 \, \text{mA} = 200 \, \text{mA} \]

### Step 4: Calculate the total current
Now, substituting \( I_g \) and \( I_s \) into the equation for total current \( I \):
\[ I = I_g + I_s = 10 \, \text{mA} + 200 \, \text{mA} = 210 \, \text{mA} \]

### Conclusion
The maximum current that can pass through the galvanometer when a shunt resistance of \( 2 \, \Omega \) is connected is \( 210 \, \text{mA} \).