In a series LR ac circuit, it is observed that at the instant voltage across the source ... what is the reactance (in Ω ) of the inductor?

Question

In a series LR circuit connected to an alternating voltage source, it is observed that at the instant voltage across the source is maximum, voltage across the inductor is 3V and voltage across the resistance is 4V. If resistance is 2Ω, what is the reactance (in Ω ) of the inductor?

Answer 1

Key Information:

1. Voltage across the inductor (\( V_L \)) = \( 3 \text{V} \),
2. Voltage across the resistor (\( V_R \)) = \( 4 \text{V} \),
3. Resistance of the resistor (\( R \)) = \( 2 \Omega \),
4. Voltage across the source (\( V_s \)) is maximum.

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Step 1: Use the phasor relationship in a series LR circuit

In a series LR circuit, the voltages across the resistor (\( V_R \)) and inductor (\( V_L \)) are not in phase:

• \( V_R \) is in phase with the current,
• \( V_L \) leads the current by \(90^\circ \).

The total voltage (\( V_s \)) is the phasor sum of \( V_R \) and \( V_L \), given by:

\( V_s = \sqrt{V_R^2 + V_L^2} \)

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Step 2: Calculate the current (\( I \)) in the circuit

From Ohm's law for the resistor:

\( V_R = I R \)

Rearrange to find \( I \):

\( I = \frac{V_R}{R} = \frac{4}{2} = 2 \, \text{A} \)

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Step 3: Use the relationship between \( V_L \) and \( X_L \)

The voltage across the inductor (\( V_L \)) is related to the inductive reactance (\( X_L \)) and current (\( I \)) as:

\( V_L = I X_L \)

Rearrange to find \( X_L \):

\( X_L = \frac{V_L}{I} \).

Substitute \( \text{V} \text{ and } I=2 A \):

\( X_L = \frac{3}{2} = 1.5 \Omega. \)

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Final Answer:

The reactance of the inductor is:

\( 1.5 Ω \).