A galvanometer has resistance `1 Omega`. Maximum deflection current through it is 0.1 A

Options

A. A) To measure a current of 1A a resistance of `(1)/(10)Omega` is put in parallel with galvanometer
B. B) To measure a current of 1A a reistance of `(1)/(9) Omega` is put in parallel with galvanometer
C. C) To measure a potential difference of 10 V a resistance of `99 Omega` is put in series with galvanometer
D. D) To measure a potential difference of 10 V a resistance of `100 Omega` is put in series with galvanometer

Answer

To solve the problem regarding the galvanometer, we need to determine the appropriate resistances to be used in parallel and series with the galvanometer for measuring a specific current and voltage.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Resistance of the galvanometer (G) = 1 Ω
- Maximum deflection current (I_g) = 0.1 A

2. **Determine the Current Measurement Setup:**
- To measure a larger current (let's say 1 A), we need to use a shunt resistor (R_s) in parallel with the galvanometer.
- The total current (I) through the circuit will be the sum of the current through the galvanometer (I_g) and the current through the shunt resistor (I_s):
\[ I = I_g + I_s \]
- For maximum deflection, we want to find R_s such that when I = 1 A, I_g = 0.1 A.

3. **Calculate the Current through the Shunt Resistor:**
- Rearranging the equation gives us:
\[ I_s = I - I_g = 1 A - 0.1 A = 0.9 A \]

4. **Using the Formula for Parallel Resistors:**
- The voltage across the galvanometer (V_g) can be expressed as:
\[ V_g = I_g \times G = 0.1 A \times 1 Ω = 0.1 V \]
- The voltage across the shunt resistor (R_s) must be the same as the voltage across the galvanometer:
\[ V_s = I_s \times R_s \]
- Setting the voltages equal gives:
\[ 0.1 V = 0.9 A \times R_s \]
- Solving for R_s:
\[ R_s = \frac{0.1 V}{0.9 A} = \frac{1}{9} Ω \approx 0.111 Ω \]

5. **Determine the Series Resistance for Voltage Measurement:**
- To measure a potential difference of 10 V, we need to add a resistor (R_series) in series with the galvanometer.
- The total voltage (V_total) across the series circuit will be:
\[ V_total = I_g \times (G + R_{series}) = 10 V \]
- Rearranging gives:
\[ 10 V = 0.1 A \times (1 Ω + R_{series}) \]
- Solving for R_series:
\[ R_{series} = \frac{10 V}{0.1 A} - 1 Ω = 100 Ω - 1 Ω = 99 Ω \]

### Final Answers:
- The shunt resistor required to measure 1 A is \( \frac{1}{9} \, \Omega \).
- The series resistor required to measure 10 V is \( 99 \, \Omega \).