Question

An inductor with inductance \(L\) is connected to an \(A C\) power source that supplies \(V_{\mathrm{emf}}=20.7 \mathrm{~V}\) at \(f=733 \mathrm{~Hz}\). If the reactance of the inductor is to be \(81.52 \Omega\), what should the value of \(L\) be?

Short answer

Question: Calculate the inductance of an inductor given a reactance of 81.52 ohms and a frequency of 733 Hz. Answer: The inductance of the inductor is approximately 0.018 H.

Step-by-step solution

Identify the given values

We are given the following values: - Voltage: \(V_{emf} = 20.7V \) - Frequency: \(f = 733 Hz \) - Reactance of the inductor: \(X_L = 81.52\Omega\)

Write down the formula for reactance of an inductor

The formula for the reactance of an inductor is given by: \(X_L = 2 \pi f L\)

Rearrange the formula to solve for the inductance \(L\)

To solve for inductance \(L\), we need to rearrange the formula. Dividing both sides by \(2 \pi f\), we get: \(L = \frac{X_L}{2 \pi f}\)

Substitute the given values into the rearranged formula

We can now substitute the given values of \(X_L\) and \(f\) into the rearranged formula: \(L = \frac{81.52\Omega}{2 \pi \cdot 733 Hz}\)

Calculate the value of the inductance \(L\)

Perform the calculation to find the inductance: \(L = \frac{81.52\Omega}{2 \pi \cdot 733 Hz} \approx 0.018 H\) So, the value of the inductance \(L\) should be approximately \(0.018 H\).