A galvanometer coil has a resistance of `50 Omega` and the meter shows full scale deflection for a current of 5 mA . This galvanometer is converted into a voltmeter of range 0 - 20 V by connecting

Options

A. `3950Omega` in series with galvanometer
B. `4050Omega` in series with galvanometer
C. `3950Omega` in parallel with galvanometer
D. `4050Omega` in parallel with galvanometer

Answer

To convert a galvanometer into a voltmeter, we need to connect a resistor in series with the galvanometer. Let's solve the problem step by step.

### Step 1: Understand the given values
- Resistance of the galvanometer, \( R_g = 50 \, \Omega \)
- Full-scale deflection current, \( I_g = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A} \)
- Desired voltage range of the voltmeter, \( V = 20 \, \text{V} \)

### Step 2: Use Ohm's Law
The total voltage across the galvanometer and the series resistor can be expressed using Ohm's Law:
\[ V = I \cdot R_{\text{total}} \]
Where \( R_{\text{total}} = R_g + R \) (the resistance of the galvanometer plus the series resistor).

### Step 3: Set up the equation
Substituting the known values into the equation:
\[ 20 \, \text{V} = (5 \times 10^{-3} \, \text{A}) \cdot (R_g + R) \]
Substituting \( R_g = 50 \, \Omega \):
\[ 20 = (5 \times 10^{-3}) \cdot (50 + R) \]

### Step 4: Solve for \( R \)
First, we can simplify the equation:
\[ 20 = 0.005 \cdot (50 + R) \]
Dividing both sides by \( 0.005 \):
\[ 4000 = 50 + R \]
Now, isolate \( R \):
\[ R = 4000 - 50 \]
\[ R = 3950 \, \Omega \]

### Step 5: Conclusion
To convert the galvanometer into a voltmeter with a range of 0 - 20 V, a resistor of \( 3950 \, \Omega \) should be connected in series with the galvanometer.

### Final Answer
The resistance to be connected in series with the galvanometer is \( R = 3950 \, \Omega \).

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