A galvanometer having resistance of `50 Omega` requires a current of `100 Omega A` to given full scale deflection. How much resistance is required to convert it into an ammeter of range of 10 A ?

Options

A. `5xx10^(-3)Omega` in series
B. `5xx10^(-4)Omega` in parallel
C. `10^(5)Omega` in series
D. `10^(5)Omega` in parallel

Answer

To solve the problem of converting a galvanometer into an ammeter with a specific range, we can follow these steps:

### Step 1: Understand the given values
- Resistance of the galvanometer (G) = 50 Ω
- Full-scale deflection current (Ig) = 100 µA = 100 × 10^(-6) A
- Desired range of the ammeter (I) = 10 A

### Step 2: Calculate the current through the shunt resistor
The current flowing through the galvanometer is 100 µA, and the total current through the ammeter will be 10 A. Therefore, the current flowing through the shunt resistor (R) can be calculated as:
\[ I_R = I - I_g = 10 A - 100 µA = 10 A - 0.0001 A = 9.9999 A \]
For practical purposes, we can approximate this to:
\[ I_R \approx 10 A \]

### Step 3: Apply the principle of parallel resistances
Since the galvanometer and the shunt resistor are in parallel, the voltage across both must be the same. The voltage across the galvanometer can be expressed as:
\[ V_g = I_g \cdot G = (100 \times 10^{-6}) \cdot 50 \]

Calculating this gives:
\[ V_g = 0.005 V = 5 mV \]

### Step 4: Set up the equation for the shunt resistor
The voltage across the shunt resistor (R) is also equal to the voltage across the galvanometer:
\[ V_R = I_R \cdot R \]
Setting the voltages equal gives:
\[ I_g \cdot G = I_R \cdot R \]
Substituting the values we have:
\[ (100 \times 10^{-6}) \cdot 50 = 9.9999 \cdot R \]

### Step 5: Solve for R
Now we can solve for R:
\[ R = \frac{(100 \times 10^{-6}) \cdot 50}{9.9999} \]
Calculating the numerator:
\[ (100 \times 10^{-6}) \cdot 50 = 0.005 \]
So,
\[ R = \frac{0.005}{9.9999} \approx 0.0005 \, \Omega \]
or
\[ R = 5 \times 10^{-4} \, \Omega \]

### Step 6: Conclusion
The resistance required to convert the galvanometer into an ammeter of range 10 A is approximately:
\[ R \approx 5 \times 10^{-4} \, \Omega \]