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, we will follow the information given in the question and apply the relevant formulas.

### Step 1: Determine the maximum current through the galvanometer (Ig)

The galvanometer gives a deflection of 5 divisions per milliampere, and there are 50 divisions on the scale.

To find the maximum current (Ig) that can pass through the galvanometer, we can use the following formula:

\[ Ig = \frac{\text{Total divisions}}{\text{Divisions per mA}} = \frac{50 \text{ divisions}}{5 \text{ divisions/mA}} = 10 \text{ mA} \]

### Step 2: Identify the resistance values

We are given:
- Resistance of the galvanometer (Rg) = 40 Ω
- Shunt resistance (Rs) = 2 Ω

### Step 3: Use the formula for maximum current (I) through the circuit with shunt resistance

The formula to calculate the maximum current (I) that can pass through the galvanometer when a shunt resistance is connected is:

\[ I = \left( \frac{Rg + Rs}{Rg} \right) \times Ig \]

### Step 4: Substitute the values into the formula

Substituting the known values into the formula:

\[ I = \left( \frac{40 \, \Omega + 2 \, \Omega}{40 \, \Omega} \right) \times 10 \times 10^{-3} \, \text{A} \]

### Step 5: Simplify the equation

Calculating the expression:

\[ I = \left( \frac{42}{40} \right) \times 10 \times 10^{-3} \]

\[ I = 1.05 \times 10 \times 10^{-3} \, \text{A} \]

\[ I = 1.05 \times 10^{-2} \, \text{A} = 0.105 \, \text{A} = 105 \, \text{mA} \]

### Step 6: Final calculation for maximum current

Now, we can calculate the maximum current that can pass through the galvanometer with the shunt resistance:

\[ I = \left( \frac{42}{40} \right) \times 10 \times 10^{-3} = 1.05 \times 10^{-2} \, \text{A} = 105 \, \text{mA} \]

### Conclusion

The maximum current that can pass through the galvanometer when a shunt resistance of 2 Ω is connected is **105 mA**.

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