A microammeter has as resistance of `100 Omega` and full scale range of `50 muA`. It can be used a voltmeter or as ahigher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations
50 V range with `10 kOmega` resistance in series
b.`10 V` range with `200 kOmega` resistance in series
c. 5mA rangw with `1Omega` resistance in parallel
`10 mA` range with `1Omega` resistance in parallel

Options

A. 50 V range with `10 kOmega` resistance in series
B. 10 V range with `200 kOmega` resistance in series.
C. 5mA range with `2 Omega` resistance in parallel
D. 10 mA range with `2Omega` resistance in parallel

Answer

To solve the problem, we need to analyze the given options for using a microammeter as a voltmeter and as an ammeter. The microammeter has a resistance of \( G = 100 \, \Omega \) and a full-scale range of \( I_G = 50 \, \mu A \).

### Step 1: Using the Microammeter as a Voltmeter

When using the microammeter as a voltmeter, we need to connect a large resistance in series with it. The formula to calculate the required series resistance \( R_V \) for a given voltage \( V \) is:

\[ R_V = \frac{V}{I_G} - G \]

#### Option A: 50 V range with \( 10 k\Omega \) resistance in series

1. **Calculate \( R_V \) for \( V = 50 \, V \)**:
\[ R_V = \frac{50}{50 \times 10^{-6}} - 100 \]
\[ R_V = \frac{50}{0.00005} - 100 = 1000000 - 100 = 999900 \, \Omega \approx 10^6 \, \Omega \]
This does not match \( 10 k\Omega \).

#### Option B: 10 V range with \( 200 k\Omega \) resistance in series

2. **Calculate \( R_V \) for \( V = 10 \, V \)**:
\[ R_V = \frac{10}{50 \times 10^{-6}} - 100 \]
\[ R_V = \frac{10}{0.00005} - 100 = 200000 - 100 = 199900 \, \Omega \approx 200 k\Omega \]
This matches \( 200 k\Omega \).

### Step 2: Using the Microammeter as an Ammeter

When using the microammeter as an ammeter, we need to connect a shunt resistance \( S \) in parallel. The formula for calculating the shunt resistance \( S \) is:

\[ S = \frac{I_G \cdot G}{I - I_G} \]

#### Option C: 5 mA range with \( 1 \Omega \) resistance in parallel

3. **Calculate \( S \) for \( I = 5 \, mA \)**:
\[ S = \frac{50 \times 10^{-6} \cdot 100}{5 \times 10^{-3} - 50 \times 10^{-6}} \]
\[ S = \frac{0.005}{0.00495} \approx 1.01 \, \Omega \approx 1 \, \Omega \]
This matches \( 1 \Omega \).

#### Option D: 10 mA range with \( 1 \Omega \) resistance in parallel

4. **Calculate \( S \) for \( I = 10 \, mA \)**:
\[ S = \frac{50 \times 10^{-6} \cdot 100}{10 \times 10^{-3} - 50 \times 10^{-6}} \]
\[ S = \frac{0.005}{0.00995} \approx 0.5 \, \Omega \]
This does not match \( 1 \Omega \).

### Conclusion

- **Correct options**:
- Option B: \( 10 V \) range with \( 200 k\Omega \) resistance in series.
- Option C: \( 5 mA \) range with \( 1 \Omega \) resistance in parallel.