An inductor stores $16\text{ J}$ of magnetic field energy and dissipates $32\text{ W}$ of thermal energy due to its resistance when an a.c. current of $2\text{ A}$ (rms) and frequency $50\text{ Hz}$ flows through it. The ratio of inductive reactance to its resistance is ________

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An inductor stores $16\text{ J}$ of magnetic field energy and dissipates $32\text{ W}$ of thermal energy due to its resistance when an a.c. current of $2\text{ A}$ (rms) and frequency $50\text{ Hz}$ flows through it. The ratio of inductive reactance to its resistance is _________. ($\pi = 3.14$)

Resistance Calculation from Power Dissipation

The thermal energy dissipated ($P_R$) in an inductor due to its resistance ($R$) is given by the formula $P_R = I_{rms}^2 R$, where $I_{rms}$ is the root-mean-square current.

Given $P_R = 32\text{ W}$ and $I_{rms} = 2\text{ A}$.
Substitute the values: $32\text{ W} = (2\text{ A})^2 \times R$.
$32 = 4 \times R$.
Solving for $R$: $R = \frac{32}{4} = 8\text{ }\Omega$.

Inductance Calculation from Average Energy Stored

The energy ($U$) stored in an inductor's magnetic field varies with current. The problem states $16\text{ J}$ is stored. This is interpreted as the average energy stored ($\langle U \rangle$) over an AC cycle.

The average energy stored in an inductor is related to the peak energy ($U_{peak}$) by $\langle U \rangle = \frac{1}{2} U_{peak}$.

The peak energy is given by $U_{peak} = \frac{1}{2} L I_{peak}^2$, where $L$ is the inductance and $I_{peak}$ is the peak current.

First, find the peak current ($I_{peak}$) from the RMS current ($I_{rms}$): $I_{peak} = I_{rms} \sqrt{2} = 2\sqrt{2}\text{ A}$.

Now, relate average energy to inductance:

$\langle U \rangle = \frac{1}{2} \left( \frac{1}{2} L I_{peak}^2 \right) = \frac{1}{4} L I_{peak}^2$.
Given $\langle U \rangle = 16\text{ J}$.
Substitute values: $16\text{ J} = \frac{1}{4} L (2\sqrt{2}\text{ A})^2$.
$16 = \frac{1}{4} L (4 \times 2) = \frac{1}{4} L (8) = 2L$.
Solving for $L$: $L = \frac{16}{2} = 8\text{ H}$.

Inductive Reactance Calculation

The inductive reactance ($X_L$) is calculated using the formula $X_L = 2\pi f L$, where $f$ is the frequency and $L$ is the inductance.

Given $f = 50\text{ Hz}$ and $L = 8\text{ H}$. Use $\pi = 3.14$.
$X_L = 2 \times 3.14 \times 50\text{ Hz} \times 8\text{ H}$.
$X_L = 6.28 \times 400$.
$X_L = 2512\text{ }\Omega$.

Ratio Calculation ($X_L / R$)

Finally, calculate the ratio of inductive reactance to resistance.

$X_L = 2512\text{ }\Omega$.
$R = 8\text{ }\Omega$.
Ratio = $\frac{X_L}{R} = \frac{2512\text{ }\Omega}{8\text{ }\Omega}$.
Ratio = $314$.

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