ISRO EC 2023 Q.79: solution and answer

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

1. The quality factor of a coil is defined as $Q=2\pi \times \frac{\text{peak energy stored}}{\text{energy dissipated per cycle}}$.

2. Model the coil as an inductance $L$ in series with its winding resistance $R$, carrying a current of peak value $I_m$ at angular frequency $\omega$. The peak stored energy is $\frac{1}{2}L{I}_m^2$, and the energy dissipated per cycle is $\frac{1}{2}{I}_m^{2}RT$ with $T=2\pi /\omega$.

3. Substituting, $$Q=2\pi \frac{\frac{1}{2}L{I}_m^2}{\frac{1}{2}{I}_m^{2}R(2\pi /\omega )}=\frac{\omega L}{R}=\frac{X_L}{R}.$$

4. Option B, $R/X_L$, is the reciprocal of $Q$ (the dissipation factor), not $Q$ itself.

5. Option C, $X_L/R^2$, has units of $1/\Omega$, so it is not dimensionless and cannot be a quality factor.

6. Option D, $X_C/X_L$, compares a capacitive reactance with an inductive one. It says nothing about the coil's own loss.

7. Therefore $Q=X_L/R$, which is option A.