A voltmeter with a resistance `50 xx 10^(3) Omega` is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is

Options

A. `9 xx 10^(6) Omega`
B. `10^(5) Omega`
C. `1.5 xx 10^(5) Omega`
D. `9 xx 10^(5) Omega`

Answer

To solve the problem of finding the additional resistance needed to increase the range of a voltmeter with a resistance of \( R_v = 50 \times 10^3 \, \Omega \) (or \( 5 \times 10^4 \, \Omega \)) to three times its original range, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- The voltmeter has a resistance \( R_v = 50 \times 10^3 \, \Omega \).
- It measures a voltage \( V \) across it when a current \( I \) flows through it.

2. **Voltage Measurement**:
- The voltage measured by the voltmeter is given by Ohm's Law:
\[ V = I \times R_v \]

3. **Increasing the Range**:
- To increase the voltage range to three times, we need the new voltage \( V' \) to be:
\[ V' = 3V \]

4. **Adding Series Resistance**:
- When an additional resistance \( R_0 \) is added in series with the voltmeter, the total resistance becomes:
\[ R_{total} = R_v + R_0 \]
- The new voltage across the voltmeter when the same current \( I \) flows is:
\[ V' = I \times (R_v + R_0) \]

5. **Setting Up the Equation**:
- Since \( V' = 3V \), we can substitute \( V \) from step 2:
\[ I \times (R_v + R_0) = 3(I \times R_v) \]

6. **Simplifying the Equation**:
- Cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ R_v + R_0 = 3R_v \]

7. **Rearranging for \( R_0 \)**:
- Rearranging the equation gives:
\[ R_0 = 3R_v - R_v = 2R_v \]

8. **Substituting the Value of \( R_v \)**:
- Now substitute \( R_v = 5 \times 10^4 \, \Omega \):
\[ R_0 = 2 \times (5 \times 10^4) = 10 \times 10^4 \, \Omega = 10^5 \, \Omega \]

### Final Answer:
The additional resistance \( R_0 \) that needs to be put in series to increase the range of the voltmeter to three times is:
\[ R_0 = 10^5 \, \Omega \]