Question

If a coil of inductance L and resistance R is connected to AC source with voltage V and angular frequency \(\omega\) and current I, then the inductance L is

• A. \(\frac{1}{\omega ^{2}}\left(\sqrt{\left(\frac{V_{rms}}{I_{rms}}\right)^{2}-R^{2}}\right)\)
• B. \(\frac{1}{\omega }\left(\sqrt{\left(\frac{V_{rms}}{I_{rms}}\right)^{2}-R^{2}}\right)\)
• C. \(\frac{1}{\omega }\left(\left(\frac{V_{rms}}{I_{rms}}\right)^{2}-R^{2}\right)\)
• D. \(\frac{1}{\omega ^{2}}\left(\sqrt{\left(\frac{V_{rms}}{I_{rms}}\right)^{3}-R^{3}}\right)\)

Hint:

Impedance of a coil is the square root of R squared plus the square of omega L; rearrange for L.

Answer 1

The Correct Option is B

Step 1: Reactance route
The inductive reactance is $X_L=\sqrt{Z^{2}-R^{2}}$ and $X_L=\omega L$.

Step 2: Why the others fail
Options with $\dfrac{1}{\omega^{2}}$ or cubes are dimensionally wrong, and the option without the root skips taking the square root of $Z^{2}-R^{2}$.

Step 3: Conclusion
Dividing $X_L$ by $\omega$ gives option (B).

Final Answer:
Option (B) gives the correct expression for L. \[ \boxed{\text{(B) }\dfrac{1}{\omega}\left(\sqrt{\left(\dfrac{V_{rms}}{I_{rms}}\right)^{2}-R^{2}}\right)} \]