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 add a series resistance to allow it to measure higher voltages. Let's solve the problem step by step.

### Step 1: Identify Given Values
- Resistance of the galvanometer, \( R_g = 50 \, \Omega \)
- Full-scale deflection current, \( I_{max} = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A} \)
- Maximum voltage range for the voltmeter, \( V_{max} = 20 \, \text{V} \)

### Step 2: Calculate the Total Resistance Needed
When converting the galvanometer into a voltmeter, the total resistance \( R_{eq} \) required can be calculated using Ohm's Law:

\[ V_{max} = I_{max} \times R_{eq} \]

Rearranging the equation gives:

\[ R_{eq} = \frac{V_{max}}{I_{max}} \]

Substituting the values:

\[ R_{eq} = \frac{20 \, \text{V}}{5 \times 10^{-3} \, \text{A}} = 4000 \, \Omega \]

### Step 3: Calculate the Series Resistance Needed
The total resistance \( R_{eq} \) is the sum of the galvanometer resistance \( R_g \) and the series resistance \( R \):

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

Substituting the known values:

\[ 4000 \, \Omega = 50 \, \Omega + R \]

Now, solve for \( R \):

\[ R = 4000 \, \Omega - 50 \, \Omega = 3950 \, \Omega \]

### Conclusion
The resistance that needs to be connected in series with the galvanometer to convert it into a voltmeter with a range of 0 - 20 V is \( 3950 \, \Omega \).

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