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 Concept:
A galvanometer is converted into a voltmeter by connecting a high resistance in series. The voltage range $V$ is given by $V = I_g(G + R)$, where $I_g$ is the full-scale deflection current and $G$ is the galvanometer resistance.

Step 2: Formula Application:
Case 1: $15 = I_g(100 + X)$. Case 2 (Double range): $30 = I_g(100 + X + 1500)$.

Step 3: Explanation:
Dividing Case 2 by Case 1: $\frac{30}{15} = \frac{I_g(1600 + X)}{I_g(100 + X)}$. $2 = \frac{1600 + X}{100 + X} \implies 200 + 2X = 1600 + X$. $X = 1400 \, \Omega$.

Step 4: Final Answer:
The value of 'X' is 1400 $\Omega$.