When a current of 5mA is passed through a galvanometer having a coil of resistance `15Omega`, it shows full sacle deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range `0-10V` is:

Options

A. `4.005xx10^(3)Omega`
B. `1.985xx10^(3)Omega`
C. `2.045xx10^(3)Omega`
D. `2.535xx10^(3)Omega`

Answer

To solve the problem, we need to determine the resistance that must be added in series with the galvanometer to convert it into a voltmeter with a range of 0-10V. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Current through the galvanometer (I) = 5 mA = 5 × 10^-3 A
- Resistance of the galvanometer (Rg) = 15 Ω
- Maximum voltage (V) = 10 V

### Step 2: Apply Ohm's Law
The total voltage across the series combination of the galvanometer and the additional resistance (R) can be expressed using Ohm's Law:
\[ V = I \times (R + Rg) \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ 10 = (5 \times 10^{-3}) \times (R + 15) \]

### Step 4: Rearranging the equation
To isolate R, we can rearrange the equation:
\[ 10 = 5 \times 10^{-3} \times R + 5 \times 10^{-3} \times 15 \]

Calculating \( 5 \times 10^{-3} \times 15 \):
\[ 5 \times 10^{-3} \times 15 = 0.075 \]

Now substituting this back:
\[ 10 = 5 \times 10^{-3} \times R + 0.075 \]

### Step 5: Isolate R
Now, we can isolate R:
\[ 10 - 0.075 = 5 \times 10^{-3} \times R \]
\[ 9.925 = 5 \times 10^{-3} \times R \]

### Step 6: Solve for R
Now divide both sides by \( 5 \times 10^{-3} \):
\[ R = \frac{9.925}{5 \times 10^{-3}} \]
\[ R = 1985 \, \Omega \]

### Step 7: Conclusion
The resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0-10V is approximately:
\[ R \approx 1985 \, \Omega \]