A galvenomenter of resistance `20 Omega` shows a deflection of 10 divisions when a currect of 1 mA is passed through it. If a shunt of `4 Omega` is concented and there are 50 divisions on the scale. The range of the gallvanometer is

Options

A. `1 A`
B. `3A`
C. `10 mA`
D. `30 mA`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
- The galvanometer has a resistance \( R_g = 20 \, \Omega \).
- It shows a deflection of 10 divisions for a current \( I_g = 1 \, \text{mA} \).
- A shunt of resistance \( R_s = 4 \, \Omega \) is connected in parallel.
- The galvanometer scale has 50 divisions.

### Step 2: Determine the maximum current for full-scale deflection
Since 10 divisions correspond to 1 mA, we can find the current that corresponds to the full scale (50 divisions):
\[ I_{\text{max}} = 5 \times I_g = 5 \times 1 \, \text{mA} = 5 \, \text{mA} \]

### Step 3: Use the formula for current in the galvanometer with shunt
When a shunt is connected in parallel with the galvanometer, the total current \( I \) flowing through the circuit can be expressed as:
\[ I = I_g + I_s \]
Where \( I_s \) is the current through the shunt. The voltage across the galvanometer and the shunt is the same, so we can use Ohm's law:
\[ I_g \cdot R_g = I_s \cdot R_s \]

### Step 4: Rearranging the equation
From the above equation, we can express \( I_s \):
\[ I_s = \frac{I_g \cdot R_g}{R_s} \]

### Step 5: Substitute \( I_g \) and solve for \( I \)
Substituting \( I_g = 5 \, \text{mA} \):
\[ I_s = \frac{5 \, \text{mA} \cdot 20 \, \Omega}{4 \, \Omega} = \frac{100 \, \text{mA} \cdot \Omega}{4 \, \Omega} = 25 \, \text{mA} \]

Now substituting \( I_s \) back into the total current equation:
\[ I = I_g + I_s = 5 \, \text{mA} + 25 \, \text{mA} = 30 \, \text{mA} \]

### Conclusion
The range of the galvanometer with the shunt connected is \( 30 \, \text{mA} \).