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

In a series circuit comprising a resistor and an ideal inductor, the total voltage across both components, equivalent to the supply voltage of \( 100 \sqrt{2} \, \text{V} \) (RMS), is distributed such that the voltage drop across the resistor and the inductor are identical. This equality in voltage drop arises from the circuit's total impedance, which mandates equal potential differences across each element.Step by Step Solution: Step 1: Designate the voltage across the resistor as \( V_R \) and the voltage across the inductor as \( V_L \). It is established that \( V_R = V_L \). Step 2: The total voltage, \( V_{\text{total}} \), is the vectorial sum of \( V_R \) and \( V_L \). Due to their phase difference (90 degrees), the Pythagorean theorem is applied:\[V_{\text{total}} = \sqrt{V_R^2 + V_L^2}\] Step 3: Substituting the condition \( V_R = V_L \) into the equation yields:\[V_{\text{total}} = \sqrt{V_R^2 + V_R^2} = \sqrt{2V_R^2} = V_R \sqrt{2}\] Step 4: Given the total voltage \( V_{\text{total}} = 100\sqrt{2} \, \text{V} \), we can determine \( V_R \):\[100\sqrt{2} = V_R \sqrt{2} \quad \Rightarrow \quad V_R = 100 \, \text{V}\]The voltmeter reading is therefore:\[\boxed{100 \, \text{V}}.\]