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, we need to determine the additional resistance required to increase the range of the voltmeter to three times its original measurement capability.

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistance of the voltmeter, \( R_v = 50 \times 10^3 \, \Omega \) (or \( 50 \, k\Omega \)).
- We want to increase the voltage range to three times the original, so the new voltage measurement will be \( 3V \).

2. **Understanding the relationship:**
- The voltage measured by the voltmeter is given by \( V = I \cdot R_v \).
- If we add an additional resistance \( R \) in series with the voltmeter, the total resistance becomes \( R_t = R_v + R \).
- The new voltage measured will be \( V' = I \cdot R_t = I \cdot (R_v + R) \).

3. **Setting up the equation:**
- Since we want the new voltage \( V' \) to be three times the original voltage \( V \), we can write:
\[ V' = 3V \]
- Substituting the expressions for \( V \) and \( V' \):
\[ I \cdot (R_v + R) = 3 \cdot (I \cdot R_v) \]

4. **Simplifying the equation:**
- We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ R_v + R = 3R_v \]
- Rearranging gives:
\[ R = 3R_v - R_v = 2R_v \]

5. **Calculating the additional resistance:**
- Now substitute \( R_v \) into the equation:
\[ R = 2 \cdot (50 \times 10^3) = 100 \times 10^3 \, \Omega \]
- Therefore, the additional resistance required is:
\[ R = 100 \, k\Omega \]

### Final Answer:
The additional resistance to be put in series is \( 100 \, k\Omega \).

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