An `80 Omega` galvanometer deflects full-scale for a potential of `20 mV`. A voltmeter deflcting full-scale of `5 V` is to be made using this galvanometer. We must connect

Options

A. a resistance of `19.92 k Omega` parallel to the galvanometer
B. a resistance of `19.92 k Omega` in series with the galvanometer
C. a resistance of `20 k Omega` parallel to the galvanometer
D. a resistance of `20 k Omega` in series with the galvanometer

Explanation

To solve the problem of converting an 80-ohm galvanometer that deflects full-scale for a potential of 20 mV into a voltmeter that deflects full-scale at 5 V, we need to determine the required shunt resistance. Here’s a step-by-step solution:

### Step 1: Determine the current through the galvanometer at full-scale deflection
The current through the galvanometer (Ig) can be calculated using Ohm's law:
\[ I_g = \frac{V_g}{R_g} \]
Where:
- \( V_g = 20 \, \text{mV} = 20 \times 10^{-3} \, \text{V} \)
- \( R_g = 80 \, \Omega \)

Substituting the values:
\[ I_g = \frac{20 \times 10^{-3}}{80} = \frac{20}{80} \times 10^{-3} = \frac{1}{4} \times 10^{-3} = 0.25 \, \text{mA} \]

### Step 2: Set up the equation for the voltmeter
When the galvanometer is converted into a voltmeter, we need to find the total resistance (R_total) in the circuit when the voltmeter reads 5 V. The total voltage across the galvanometer and the shunt resistor (S) is equal to the voltmeter reading:
\[ V_{AB} = I_g \times (R_g + S) \]
Where:
- \( V_{AB} = 5 \, \text{V} \)

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ 5 = I_g \times (R_g + S) \]
Substituting \( I_g = 0.25 \times 10^{-3} \, \text{A} \) and \( R_g = 80 \, \Omega \):
\[ 5 = 0.25 \times 10^{-3} \times (80 + S) \]

### Step 4: Solve for S
Rearranging the equation gives:
\[ 80 + S = \frac{5}{0.25 \times 10^{-3}} \]
Calculating the right side:
\[ 80 + S = \frac{5}{0.00025} = 20000 \]
Now, solving for S:
\[ S = 20000 - 80 = 19920 \, \Omega \]

### Step 5: Convert to kilo-ohms
To express S in kilo-ohms:
\[ S = 19920 \, \Omega = 19.92 \, \text{k}\Omega \]

### Conclusion
The required shunt resistance to convert the galvanometer into a voltmeter that reads 5 V at full-scale deflection is:
\[ S = 19.92 \, \text{k}\Omega \]