A galvanometer of resistance `25Omega` measures `10^(-3)A` . shunt required to increase range up tow 2 A is

Options

A. `12.5 mOmega`
B. `0.125 mOmega`
C. `0.125 Omega`
D. `1.25 mOmega`

Answer

To solve the problem of finding the required shunt resistance for a galvanometer, we can follow these steps:

### Step 1: Understand the given values
- Resistance of the galvanometer (G) = 25 Ω
- Current through the galvanometer (Ig) = \(10^{-3}\) A = 0.001 A
- Desired maximum current (I) = 2 A

### Step 2: Set up the relationship
When a shunt resistor (S) is connected in parallel with the galvanometer, the total current (I) splits into two parts:
- Current through the galvanometer (Ig)
- Current through the shunt (Is)

From the current division in parallel circuits, we have:
\[ I = Ig + Is \]

### Step 3: Relate the voltages across the galvanometer and shunt
Since the galvanometer and the shunt are in parallel, the voltage across both must be the same:
\[ V_g = V_s \]
Using Ohm's law, we can express the voltages as:
\[ Ig \cdot G = Is \cdot S \]

### Step 4: Express Is in terms of I and Ig
From the first equation, we can express Is:
\[ Is = I - Ig \]
Substituting this into the voltage equation gives:
\[ Ig \cdot G = (I - Ig) \cdot S \]

### Step 5: Substitute the known values
Now, substituting the known values into the equation:
\[ 0.001 \cdot 25 = (2 - 0.001) \cdot S \]

### Step 6: Solve for S
Calculating the left side:
\[ 0.025 = (1.999) \cdot S \]
Now, we can solve for S:
\[ S = \frac{0.025}{1.999} \]

### Step 7: Calculate S
Calculating the value:
\[ S \approx 0.012506 \, \text{Ω} \]
This can be converted to milli-ohms:
\[ S \approx 12.5 \, \text{mΩ} \]

### Final Answer
The required shunt resistance is approximately **12.5 mΩ**.

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