To solve the problem, we need to find the resistances required to convert a galvanometer into a voltmeter and an ammeter.
### Given Data:
- Resistance of galvanometer (R_g) = 100 Ω
- Full-scale deflection current (I_g) = 50 µA = 50 × 10^(-6) A
- Voltage range for voltmeter = 50 V
- Current range for ammeter = 10 mA = 10 × 10^(-3) A
### Part (a): To convert the galvanometer into a voltmeter of range 50 V
1. **Determine the total current (I)**:
The full-scale deflection current for the galvanometer is 50 µA. When using the galvanometer as a voltmeter, the total current flowing through the circuit will be the same as the galvanometer's full-scale deflection current.
\[
I = I_g = 50 \times 10^{-6} \, \text{A}
\]
2. **Use Ohm's Law to find the total resistance (R_total)**:
The voltage across the galvanometer and the series resistance (R) when the full-scale deflection is reached is given by:
\[
V = I \cdot R_{total}
\]
Where \( R_{total} = R_g + R \). Thus,
\[
50 = (50 \times 10^{-6}) \cdot (100 + R)
\]
3. **Rearranging the equation**:
\[
50 = 50 \times 10^{-6} \cdot (100 + R)
\]
Dividing both sides by \( 50 \times 10^{-6} \):
\[
1,000,000 = 100 + R
\]
4. **Solve for R**:
\[
R = 1,000,000 - 100 = 999,900 \, \Omega
\]
Thus, for part (a), the resistance required to convert the galvanometer into a voltmeter of range 50 V is approximately:
\[
R \approx 10^6 \, \Omega
\]
### Part (b): To convert the galvanometer into an ammeter of range 10 mA
1. **Determine the total current (I)**:
The total current for the ammeter is given as 10 mA.
\[
I = 10 \times 10^{-3} \, \text{A}
\]
2. **Use the formula for parallel resistances**:
When using the galvanometer as an ammeter, we will connect a shunt resistance (R_s) in parallel with the galvanometer. The voltage across both components will be the same.
\[
I = I_g + I_s
\]
Where \( I_s \) is the current through the shunt resistance. Rearranging gives:
\[
I_s = I - I_g = 10 \times 10^{-3} - 50 \times 10^{-6}
\]
3. **Calculate I_s**:
\[
I_s \approx 10 \times 10^{-3} - 0.00005 \approx 0.00995 \, \text{A}
\]
4. **Using the voltage across the galvanometer**:
The voltage across the galvanometer can be expressed as:
\[
V_g = I_g \cdot R_g = (50 \times 10^{-6}) \cdot 100 = 0.005 \, \text{V}
\]
5. **Setting up the equation for the shunt resistance**:
The voltage across the shunt resistance is the same:
\[
V_g = I_s \cdot R_s
\]
Thus,
\[
0.005 = (0.00995) \cdot R_s
\]
6. **Solving for R_s**:
\[
R_s = \frac{0.005}{0.00995} \approx 0.5025 \, \Omega
\]
Therefore, for part (b), the resistance required to convert the galvanometer into an ammeter of range 10 mA is approximately:
\[
R_s \approx 0.5 \, \Omega
\]
### Summary of Results:
- (a) Resistance for voltmeter: \( R \approx 10^6 \, \Omega \)
- (b) Resistance for ammeter: \( R_s \approx 0.5 \, \Omega \)
