Resistance of a Galvanometer coil is `8Omega` and `2Omega` Shunt resistance is connected with it. If main current is 1 A then the current flow through `2Omega` resistance will be :

Options

A. 0.2 A
B. 0.8 A
C. 0.1 A
D. 0.4 A

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Circuit Configuration
The galvanometer and the shunt resistor are connected in parallel. The total current (I) entering the parallel combination is given as 1 A.

### Step 2: Define Variables
Let:
- \( I_1 \) = current through the galvanometer (8 Ω)
- \( I_2 \) = current through the shunt resistor (2 Ω)
- Total current \( I = I_1 + I_2 = 1 \, A \)

### Step 3: Apply Ohm's Law
Since the galvanometer and the shunt are in parallel, the voltage across both must be the same. Using Ohm's law (V = I × R), we can express the voltages as:
- Voltage across the galvanometer: \( V = I_1 \times 8 \)
- Voltage across the shunt: \( V = I_2 \times 2 \)

Setting these equal gives us:
\[ I_1 \times 8 = I_2 \times 2 \]

### Step 4: Rearrange the Equation
From the equation \( I_1 \times 8 = I_2 \times 2 \), we can express \( I_2 \) in terms of \( I_1 \):
\[ I_2 = 4 \times I_1 \]

### Step 5: Substitute into Total Current Equation
Now substitute \( I_2 \) back into the total current equation:
\[ I = I_1 + I_2 \]
\[ 1 = I_1 + 4 \times I_1 \]
\[ 1 = 5 \times I_1 \]

### Step 6: Solve for \( I_1 \)
Now, solve for \( I_1 \):
\[ I_1 = \frac{1}{5} = 0.2 \, A \]

### Step 7: Calculate \( I_2 \)
Now substitute \( I_1 \) back into the equation for \( I_2 \):
\[ I_2 = 4 \times I_1 = 4 \times 0.2 = 0.8 \, A \]

### Conclusion
The current flowing through the 2 Ω shunt resistance is \( 0.8 \, A \).

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