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 that should be added in series with the galvanometer. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Resistance of the galvanometer, \( R_g = 50 \, \Omega \)
- Full-scale deflection current, \( I_g = 10 \, \text{mA} = 0.01 \, \text{A} \)
- Desired voltage range of the voltmeter, \( V = 100 \, \text{V} \)

### Step 2: Use Ohm's Law
When the galvanometer is converted into a voltmeter, the total voltage across the series combination of the galvanometer and the additional resistance \( R \) must equal the desired voltage. According to Ohm's Law:

\[ V = I \cdot R_{\text{eq}} \]

where \( R_{\text{eq}} \) is the equivalent resistance of the galvanometer and the added resistor, and \( I \) is the current through the circuit when the voltmeter is at full scale deflection.

### Step 3: Calculate the equivalent resistance
The equivalent resistance \( R_{\text{eq}} \) is given by:

\[ R_{\text{eq}} = R_g + R \]

### Step 4: Set up the equation using the desired voltage
At full-scale deflection, the current through the galvanometer is \( I_g = 0.01 \, \text{A} \). Therefore, we can write:

\[ 100 = 0.01 \cdot (50 + R) \]

### Step 5: Solve for \( R \)
First, simplify the equation:

\[ 100 = 0.01 \cdot (50 + R) \]
\[ 100 = 0.5 + 0.01R \]

Now, isolate \( R \):

\[ 100 - 0.5 = 0.01R \]
\[ 99.5 = 0.01R \]

Now, divide both sides by \( 0.01 \):

\[ R = \frac{99.5}{0.01} = 9950 \, \Omega \]

### Step 6: Conclusion
The resistance that should be added in series with the galvanometer to convert it into a voltmeter of range 100 V is:

\[ R = 9950 \, \Omega \]