Question

A voltmeter with internal resistance of \(x\) Ω can be used to measure upto 20 V. In order to increase its measuring range to 30 V, the required modification is to ________.

• A. connect resistor of \(\frac{x}{2}\) Ω, in series with voltmeter.
• B. connect resistor of \(\frac{x}{2}\) Ω, in parallel to voltmeter.
• C. connect a resistor of \(x\) Ω in series with voltmeter.
• D. connect resistor of \(2x\) Ω in parallel to voltmeter.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A voltmeter is essentially a galvanometer with a high resistance in series. To increase the range of a voltmeter, an additional high resistance must be connected in series so that the device can withstand higher voltages without exceeding its full-scale deflection current.
Step 2: Key Formula or Approach:
The current required for full-scale deflection of the original voltmeter is \(I_g = \frac{V_{\text{old}}}{R_{\text{internal}}}\).
The new total resistance required for the upgraded voltmeter to measure \(V_{\text{new}}\) is \(R_{\text{total}} = \frac{V_{\text{new}}}{I_g}\).
The additional series resistor required is \(R_{\text{series}} = R_{\text{total}} - R_{\text{internal}}\).
Step 3: Detailed Explanation:
Original parameters:
Internal resistance = \(x \, \Omega\).
Original range \(V_{\text{old}} = 20 \text{ V}\).
Full-scale deflection current \(I_g = \frac{20}{x} \text{ A}\).
New desired parameters:
New range \(V_{\text{new}} = 30 \text{ V}\).
New total resistance \(R_{\text{total}} = \frac{30}{I_g} = \frac{30}{\frac{20}{x}} = \frac{30x}{20} = 1.5x \, \Omega\).
The additional resistance needed in series is:
\(R_{\text{series}} = R_{\text{total}} - R_{\text{internal}} = 1.5x - x = 0.5x = \frac{x}{2} \, \Omega\).
Step 4: Final Answer:
We must connect a resistor of \(\frac{x}{2} \, \Omega\) in series with the voltmeter.