A galvanometer with a scale divided into `100` equal divisions has a current sensitivity of `10` divisions per mA and a voltage sensitivity of `2` divisions per mV. What adoptions are required to read (i) `5A` for full scale and (ii) 1 division per volt?

Options

A. `4Omega`
B. `5Omega`
C. `3Omega`
D. `7Omega`

Answer

To solve the problem, we need to determine the necessary adaptations to read (i) 5A for full scale and (ii) 1 division per volt using the given galvanometer specifications.

### Given Data:
- Total divisions of galvanometer = 100
- Current sensitivity = 10 divisions/mA
- Voltage sensitivity = 2 divisions/mV

### Step 1: Determine the maximum current for full scale deflection
The galvanometer shows full scale deflection (100 divisions) at a current of:
\[ \text{Current at full scale} = \frac{100 \text{ divisions}}{10 \text{ divisions/mA}} = 10 \text{ mA} = 0.01 \text{ A} \]

### Step 2: Calculate the required shunt resistance for 5A
To convert the galvanometer into an ammeter that can read up to 5A, we need to use a shunt resistor (R_s). The formula for the shunt resistance is:
\[ R_s = \frac{I_g}{I - I_g} R_g \]
where:
- \(I_g\) = maximum current through the galvanometer (0.01 A)
- \(I\) = total current (5 A)
- \(R_g\) = resistance of the galvanometer

First, we need to find \(R_g\) (the resistance of the galvanometer). The voltage across the galvanometer at full scale deflection (100 divisions) is calculated as follows:
\[ \text{Voltage at full scale} = \frac{100 \text{ divisions}}{2 \text{ divisions/mV}} = 50 \text{ mV} = 0.05 \text{ V} \]
Now, using Ohm's law to find \(R_g\):
\[ R_g = \frac{V}{I_g} = \frac{0.05 \text{ V}}{0.01 \text{ A}} = 5 \, \Omega \]

Now substituting the values into the shunt resistance formula:
\[ R_s = \frac{0.01 \text{ A}}{5 \text{ A} - 0.01 \text{ A}} \times 5 \, \Omega = \frac{0.01}{4.99} \times 5 \approx 0.01002 \, \Omega \]

### Step 3: Calculate the required series resistance for 1 division per volt
To convert the galvanometer into a voltmeter that reads 1 division per volt, we need to determine the total voltage that corresponds to 100 divisions:
\[ \text{Total voltage} = 100 \text{ divisions} \times 0.5 \text{ mV/division} = 50 \text{ mV} = 0.05 \text{ V} \]
For 1 division per volt, we need to scale this up to 100V:
\[ \text{Voltage across galvanometer} = 100 \text{ V} \]
Now, using the formula for the series resistance \(R_s\) for the voltmeter:
\[ R_s = \frac{V - I_g R_g}{I_g} \]
Substituting the values:
\[ R_s = \frac{100 \text{ V} - 0.01 \text{ A} \times 5 \, \Omega}{0.01 \text{ A}} = \frac{100 - 0.05}{0.01} = \frac{99.95}{0.01} = 9995 \, \Omega \]

### Summary of Results:
1. **Shunt resistance for ammeter (5A)**: \(R_s \approx 0.01002 \, \Omega\)
2. **Series resistance for voltmeter (1 division per volt)**: \(R_s = 9995 \, \Omega\)