A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be

Options

A. 0.001
B. 0.01
C. 1
D. 0.05

Answer

To convert a millivolt meter into an ammeter, we need to calculate the value of the shunt resistor required. Here’s a step-by-step solution:

### Step 1: Understand the parameters
We have a millivolt meter with a range of 25 millivolts (mV) and we want to convert it into an ammeter with a range of 25 amperes (A).

### Step 2: Define the variables
- Let \( I \) be the total current (25 A).
- Let \( I_G \) be the current through the millivolt meter.
- Let \( I_S \) be the current through the shunt resistor.
- The voltage across the millivolt meter, \( V_G \), is 25 mV.

### Step 3: Relate the currents
The total current \( I \) is the sum of the current through the millivolt meter and the current through the shunt resistor:
\[ I = I_G + I_S \]
Thus, we can express the current through the shunt resistor as:
\[ I_S = I - I_G = 25 - I_G \]

### Step 4: Apply Ohm's Law
The voltage across the shunt resistor \( V_S \) is equal to the voltage across the millivolt meter:
\[ V_G = V_S = 25 \text{ mV} = 25 \times 10^{-3} \text{ V} \]

Using Ohm's Law, we can express the voltage across the shunt resistor as:
\[ V_S = S \cdot I_S \]
Where \( S \) is the shunt resistance.

### Step 5: Set up the equation
Since the voltage across the shunt resistor is the same as that across the millivolt meter, we can set up the equation:
\[ 25 \times 10^{-3} = S \cdot (25 - I_G) \]

### Step 6: Neglect \( I_G \)
Given that \( I_G \) is very small compared to 25 A, we can neglect \( I_G \) in the equation:
\[ 25 \times 10^{-3} = S \cdot 25 \]

### Step 7: Solve for \( S \)
Rearranging the equation gives:
\[ S = \frac{25 \times 10^{-3}}{25} \]
\[ S = 1 \times 10^{-3} \text{ ohms} = 0.001 \text{ ohms} \]

### Final Answer
The value of the necessary shunt resistor is:
\[ S = 0.001 \text{ ohms} \text{ or } 1 \text{ milliohm} \]