A galvanometer having `30` divisions has a current sensitivity of `20muA//di vision`. It has a resistance of `25Omega`. How will you convert it into an ammeter upto 1 ampere? How will you convert this ammeter into a voltmeter up to 1 volt?

Options

A. For B to have the greatest speed, choose `m_(B) = m_(A)`
B. For B to have the greatest momentum, choose `m_(B) lt lt m_(A)`
C. For B to have the greatest speed, choose `m_(B) lt lt M_(A)`
D. For the maximum fraction of kinetic energy transfer, choose `m_(B) = m_(A)`

Explanation

To convert the given galvanometer into an ammeter and then into a voltmeter, we can follow these steps:

### Step 1: Determine the Full Scale Deflection Current (Ig)
The full-scale deflection current \( I_g \) can be calculated using the formula:
\[ I_g = n \times I_s \]
where:
- \( n = 30 \) (number of divisions)
- \( I_s = 20 \, \mu A = 20 \times 10^{-6} \, A \)

Calculating \( I_g \):
\[ I_g = 30 \times 20 \times 10^{-6} = 600 \times 10^{-6} = 6 \times 10^{-4} \, A \]

### Step 2: Calculate the Shunt Resistance (S) for the Ammeter
To convert the galvanometer into an ammeter, we need to find the shunt resistance \( S \) using the formula:
\[ S = \frac{I_g \times G}{I - I_g} \]
where:
- \( G = 25 \, \Omega \) (resistance of the galvanometer)
- \( I = 1 \, A \) (maximum current for the ammeter)

Substituting the values:
\[ S = \frac{6 \times 10^{-4} \times 25}{1 - 6 \times 10^{-4}} = \frac{150 \times 10^{-4}}{0.9994} \]

Calculating \( S \):
\[ S \approx 0.015 \, \Omega \]

### Step 3: Calculate the Effective Resistance of the Ammeter (G')
The effective resistance \( G' \) of the ammeter can be calculated using:
\[ G' = \frac{G \times S}{G + S} \]
Substituting the values:
\[ G' = \frac{25 \times 0.015}{25 + 0.015} = \frac{0.375}{25.015} \]

Calculating \( G' \):
\[ G' \approx 0.01499 \, \Omega \]

### Step 4: Convert the Ammeter into a Voltmeter
To convert the ammeter into a voltmeter, we need to find the series resistance \( R \) using:
\[ R = \frac{V}{I_g} - G \]
where:
- \( V = 1 \, V \) (maximum voltage for the voltmeter)

Substituting the values:
\[ R = \frac{1}{6 \times 10^{-4}} - 25 \]

Calculating \( R \):
\[ R = 1666.67 - 25 \approx 1641.67 \, \Omega \]

### Final Results
- The shunt resistance \( S \) for the ammeter is approximately \( 0.015 \, \Omega \).
- The series resistance \( R \) for the voltmeter is approximately \( 1641.67 \, \Omega \).