Question

An inductor stores \(16\,\text{J}\) of magnetic field energy and dissipates \(32\,\text{W}\) of thermal energy due to its resistance when an alternating current of \(2\,\text{A}\) (rms) and frequency \(50\,\text{Hz}\) flows through it. The ratio of inductive reactance to resistance is _______. \((\pi=3.14)\) Given: \[ U=16\,\text{J},\quad P=32\,\text{W},\quad I=2\,\text{A},\quad f=50\,\text{Hz} \]

Hint:

Always calculate \(L\) and \(R\) separately using energy and power relations before finding reactance ratios.

Answer 1

Correct Answer: 314

Given data includes:
Magnetic field energy, \(U = 16 \, \text{J}\)
Power dissipated, \(P = 32 \, \text{W}\)
Current, \(I = 2 \, \text{A}\)
Frequency, \(f = 50 \, \text{Hz}\)
We need to find the ratio of inductive reactance \((X_L)\) to resistance \((R)\).
First, solve for \(L\) using the energy equation:
\[U = \frac{1}{2} L I^2\]
\[L = \frac{2U}{I^2} = \frac{2 \times 16}{4} = 8 \, \text{H}\]
Inductive reactance is given by:
\[X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 8 = 2512 \, \Omega\]
Use the power dissipated to find the resistance:
\[P = I^2 R\]
\[R = \frac{P}{I^2} = \frac{32}{4} = 8 \, \Omega\]
Finally, the ratio \(\frac{X_L}{R}\) is:
\[\frac{X_L}{R} = \frac{2512}{8} = 314\]
This value satisfies the given range 314,314.