Question

A galvanometer of resistance 100 \(\Omega\) gives a full scale deflection for a current of 1 mA through it. The resistance required to convert it into a voltmeter which can read up to 2 V is:

• A. 1175 \(\Omega\)
• B. 1200 \(\Omega\)
• C. 1525 \(\Omega\)
• D. 1900 \(\Omega\)
• E. 2025 \(\Omega\)

Hint:

To convert a galvanometer into a voltmeter, add a series resistance so that the voltage drop across the combination is equal to the maximum voltage that the voltmeter can measure.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
To convert a galvanometer into a voltmeter, a high resistance (called a series resistance or multiplier) must be connected in series with the galvanometer. This limits the current to the full-scale deflection current ($I_g$) when the maximum desired voltage ($V$) is applied.
Step 2: Key Formula or Approach
The total resistance of the voltmeter is $R_{total} = G + R$, where $G$ is the galvanometer resistance and $R$ is the series resistance. According to Ohm's Law: \[ V = I_g(G + R) \] Rearranging for $R$: \[ R = \frac{V}{I_g} - G \]
Step 3: Detailed Calculation
1. Identify the values: - $V = 2$ V - $I_g = 1$ mA = $10^{-3}$ A - $G = 100$ Ω 2. Apply the formula: - $R = \frac{2}{10^{-3}} - 100$ - $R = 2000 - 100$ - $R = 1900$ Ω
Step 4: Final Answer
The resistance required is 1900 Ω.