Question

With a resistance 'X' connected in series with a galvanometer of resistance 100$\Omega$, it acts as a voltmeter of range 0 – 15 V. To double the range, a resistance of 1500$\Omega$ is to be connected in series with 'X'. The value of 'X' in ohm is \dots

• A. 900
• B. 1100
• C. 1400
• D. 1600

Hint:

If you want to increase the range of a voltmeter by a factor of $n$ (here $n=2$), the total new resistance of the entire instrument must also increase by exactly that same factor $n$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A galvanometer is converted into a voltmeter by placing a high resistance in series. We are given two different ranges and the corresponding changes made to the series resistance. We must find the initial series resistance $X$.

Step 2: Key Formula or Approach:
The voltage range $V$ of a converted galvanometer is given by Ohm's Law:
$$V = I_g (G + R_s)$$
Where $I_g$ is the full-scale deflection current, $G$ is galvanometer resistance, and $R_s$ is the total series resistance.

Step 3: Detailed Explanation:
Given $G = 100 \, \Omega$.
Case 1: Range $V_1 = 15 \text{ V}$, Series resistance is $X$.
$$15 = I_g (100 + X) \quad \text{--- (Equation 1)}$$
Case 2: Range is doubled, so $V_2 = 30 \text{ V}$. An additional $1500 \, \Omega$ is added in series, making total series resistance $= X + 1500$.
$$30 = I_g (100 + X + 1500)$$
$$30 = I_g (X + 1600) \quad \text{--- (Equation 2)}$$
Divide Equation 2 by Equation 1 to cancel out the unknown current $I_g$:
$$\frac{30}{15} = \frac{I_g (X + 1600)}{I_g (100 + X)}$$
$$2 = \frac{X + 1600}{X + 100}$$
Cross-multiply to solve for $X$:
$$2(X + 100) = X + 1600$$
$$2X + 200 = X + 1600$$
$$2X - X = 1600 - 200$$
$$X = 1400 \, \Omega$$

Step 4: Final Answer:
The value of X is 1400 $\Omega$, matching option (c).