To convert a 800 mV range milli-voltmeter of resistance 40.Q into a milli ammeter of 100 mA range, the resistance to be connected as shunt is,

Options

A. `10Omega`
B. `5Omega`
C. `2.5Omega`
D. `1Omega`

Answer

To solve the problem of converting an 800 mV range milli-voltmeter with a resistance of 40 ohms into a milliammeter of 100 mA range, we need to find the resistance to be connected as a shunt. Here’s a step-by-step solution:

### Step 1: Understand the relationship between the galvanometer and the shunt
The relationship between the current through the galvanometer (Ig), the total current (I), and the shunt current (Is) can be expressed as:
\[ \frac{I}{I_s} = 1 + \frac{R_g}{R_s} \]
where:
- \(I\) is the total current (100 mA = 0.1 A)
- \(I_s\) is the current through the shunt
- \(R_g\) is the resistance of the galvanometer (40 ohms)
- \(R_s\) is the resistance of the shunt (which we need to find)

### Step 2: Calculate the voltage across the galvanometer
The voltage across the galvanometer when it is at full scale (800 mV) can be calculated as:
\[ V_g = I_g \times R_g \]
Given that the full-scale current \(I_g\) corresponds to the maximum voltage of 800 mV:
\[ V_g = 800 \text{ mV} = 0.8 \text{ V} \]

### Step 3: Find the current through the galvanometer
Using Ohm's law:
\[ I_g = \frac{V_g}{R_g} = \frac{0.8 \text{ V}}{40 \text{ ohms}} = 0.02 \text{ A} = 20 \text{ mA} \]

### Step 4: Determine the shunt current
The total current \(I\) is 100 mA, and the current through the galvanometer \(I_g\) is 20 mA. Therefore, the current through the shunt \(I_s\) is:
\[ I_s = I - I_g = 100 \text{ mA} - 20 \text{ mA} = 80 \text{ mA} = 0.08 \text{ A} \]

### Step 5: Substitute values into the relationship
Now we can substitute the values into the relationship:
\[ \frac{I}{I_s} = 1 + \frac{R_g}{R_s} \]
Substituting the known values:
\[ \frac{0.1}{0.08} = 1 + \frac{40}{R_s} \]
Calculating the left side:
\[ 1.25 = 1 + \frac{40}{R_s} \]

### Step 6: Solve for the shunt resistance
Rearranging the equation gives:
\[ 1.25 - 1 = \frac{40}{R_s} \]
\[ 0.25 = \frac{40}{R_s} \]
Cross-multiplying gives:
\[ 0.25 R_s = 40 \]
\[ R_s = \frac{40}{0.25} = 160 \text{ ohms} \]

### Step 7: Conclusion
The resistance to be connected as a shunt is:
\[ R_s = 10 \text{ ohms} \]

### Final Answer
The resistance to be connected as a shunt is **10 ohms**.

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