Question

A coil of negligible resistance is connected in series with \(90 \, \Omega\) resistor across \(120 \, \text{V}, \, 60 \, \text{Hz}\) supply. A voltmeter reads \(36 \, \text{V}\) across resistance. Inductance of the coil is:

• A. \(0.76 \, \text{H}\)
• B. \(2.86 \, \text{H}\)
• C. \(0.286 \, \text{H}\)
• D. \(0.91 \, \text{H}\)

Answer 1

The Correct Option is A

Given: \( 36 = I_{\text{rms}} R \)

Step 1: Substitute \( I_{\text{rms}} \)
\( 36 = \frac{120}{\sqrt{X_L^2 + R^2}} \times R \)
With \( R = 90 \, \Omega \), the equation becomes:
\( 36 = \frac{120 \times 90}{\sqrt{X_L^2 + 90^2}} \)

Step 2: Solve for \( X_L \)
Rearranging yields:
\( \sqrt{X_L^2 + 90^2} = 300 \)
Squaring both sides gives:
\( X_L^2 + 90^2 = 300^2 \)
\( X_L^2 = 90000 - 8100 = 81900 \)
Thus, \( X_L = 286.18 \, \Omega \)

Step 3: Calculate \( L \) using \( X_L = \omega L \)
\( \omega L = 286.18 \)
\( L = \frac{286.18}{376.8} \)
\( L = 0.76 \, \text{H} \)

Final Answer:
\( \boxed{L = 0.76 \, \text{H}} \)