A galvanometer of resistance `50Omega` given full-scale deflection for a current of 10 mA is to be changed into a voltmeter of rang 100 V . What should be the Value of resistance to be added in series with this galvanometer ?

Options

A. `9950Omega`
B. `10025Omega`
C. `10000Omega`
D. `9975Omega`

Answer

To solve the problem of converting a galvanometer into a voltmeter, we need to determine the resistance to be added in series with the galvanometer. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Resistance of the galvanometer (G) = 50 Ω
- Full-scale deflection current (Ig) = 10 mA = 0.01 A
- Desired voltage range (V) = 100 V

### Step 2: Use Ohm's Law to find the total resistance needed
The total voltage across the series combination of the galvanometer and the added resistance (R) can be expressed using Ohm's Law:
\[ V = I \times R_{\text{total}} \]
Where \( R_{\text{total}} = R + G \).

### Step 3: Set up the equation
Substituting the known values into the equation:
\[ 100 V = 0.01 A \times (R + 50 \, \Omega) \]

### Step 4: Rearrange the equation to solve for R
Rearranging gives:
\[ R + 50 = \frac{100}{0.01} \]
\[ R + 50 = 10000 \]

### Step 5: Isolate R
Now, isolate R:
\[ R = 10000 - 50 \]
\[ R = 9950 \, \Omega \]

### Conclusion
The resistance to be added in series with the galvanometer is **9950 Ω**.

---