Question

A resistor and an ideal inductor are connected in series to a \( 100 \sqrt{2} \, \text{V}, 50 \, \text{Hz} \) AC source. When a voltmeter is connected across the resistor or the inductor, it shows the same reading. The reading of the voltmeter is:

• A. \( 100 \sqrt{2} \, \text{V} \)
• B. \( 100 \, \text{V} \)
• C. \( 50 \sqrt{2} \, \text{V} \)
• D. \( 50 \, \text{V} \)

Hint:

In a series L-R circuit, the voltage across each component is the same if they share the total supply voltage.

Answer 1

The Correct Option is B

When a resistor and an ideal inductor are connected in series, the total voltage across both components is the supply voltage \( 100 \sqrt{2} \, \text{V} \), which is the RMS value. The voltmeter reading across the resistor and the inductor is the same because the voltage drop across each component will be equal. This happens because the total impedance of the circuit is such that both the resistor and the inductor have the same voltage drop. Step by Step Solution: Step 1: Let the voltage across the resistor be \( V_R \) and the voltage across the inductor be \( V_L \). Given that \( V_R = V_L \). Step 2: The total voltage \( V_{\text{total}} \) is given by the vector sum of \( V_R \) and \( V_L \). Since they are in quadrature, we use the Pythagorean theorem: \[ V_{\text{total}} = \sqrt{V_R^2 + V_L^2} \] Step 3: Substitute \( V_R = V_L \) into the equation: \[ V_{\text{total}} = \sqrt{V_R^2 + V_R^2} = \sqrt{2V_R^2} = V_R \sqrt{2} \] Step 4: Given that \( V_{\text{total}} = 100\sqrt{2} \, \text{V} \), we can solve for \( V_R \): \[ 100\sqrt{2} = V_R \sqrt{2} \quad \Rightarrow \quad V_R = 100 \, \text{V} \] Thus, the reading of the voltmeter is: \[ \boxed{100 \, \text{V}}. \]