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Step-by-Step Solution
Step 1: Understand the problem
An inductor is connected to an AC source of frequency 1000 Hz, and it has a given inductive reactance (XL) of 100 Ω. The applied voltage leads the current by 45°. We need to find the self-inductance L of the coil.
Step 2: Recall key relationships
1. Inductive reactance is given by
$X_L = \omega L = 2\pi f \, L.$
2. In an L-R circuit, the phase angle $ \phi $ between voltage and current is such that
$\tan \phi = \frac{X_L}{R}.$
Step 3: Use the given phase angle of 45°
Since the voltage leads the current by 45°, we have
$\tan 45^\circ = 1 = \frac{X_L}{R}.$
Thus,
$R = X_L.$
Step 4: Use the given impedance relationship
The total impedance $Z$ of an L-R circuit is
$Z = \sqrt{R^2 + X_L^2}.$
Given the coil’s reactance is 100 Ω when the voltage leads by 45°, the measured impedance must be 100 Ω. Hence:
$Z = 100 = \sqrt{R^2 + X_L^2}.$
Since $R = X_L$, let $R = X_L = x.$ Then,
$Z = \sqrt{x^2 + x^2} = \sqrt{2} \, x.$
Hence,
$\sqrt{2} \, x = 100 \ \Longrightarrow \ x = \frac{100}{\sqrt{2}}.$
Therefore,
$X_L = \frac{100}{\sqrt{2}}.$
Step 5: Substitute into the inductive reactance formula
We know
$X_L = 2\pi f \, L.$
Thus,
$\frac{100}{\sqrt{2}} = 2\pi \cdot 1000 \cdot L.$
So,
$L = \frac{100/\sqrt{2}}{2\pi \times 1000}.$
Step 6: Simplify the expression for L
$L = \frac{100}{\sqrt{2} \times 2\pi \times 1000}
= \frac{100}{\sqrt{2} \times 2 \times 3.14 \times 1000}
\approx 1.1 \times 10^{-2}\,\text{H}.$
Answer
The self-inductance of the coil is
$1.1 \times 10^{-2}\,\text{H}.$
