A 4.00 -mH inductor is connected to an ac voltage source of \(151.0 \mathrm{V}\) rms. If the rms current in the circuit is \(0.820 \mathrm{A},\) what is the frequency of the source?

Short Answer

Expert verified
Answer: The frequency of the AC voltage source is approximately 7.31 kHz.

Step by step solution

01

Find the inductive reactance (XL)

We can use the Ohm's law for an inductor to find the inductive reactance (\(X_L\)) by dividing the RMS voltage (\(V_{rms}\)) by the RMS current (\(I_{rms}\)): \(X_L = \frac{V_{rms}}{I_{rms}}\) Plug in the given values: \(X_L = \frac{151.0 \mathrm{V}}{0.820 \mathrm{A}} \approx 184.15 \Omega\)
02

Use the formula to find the frequency (f)

Now that we have the inductive reactance, we can use the formula that relates inductive reactance, inductance, and frequency: \(X_L = 2\pi f L\) Where \(f\) is the frequency, and \(L\) is the inductance. We can rearrange the formula to solve for the frequency: \(f = \frac{X_L}{2\pi L}\) Plug in the given values and the calculated value for \(X_L\): \(f = \frac{184.15 \Omega}{2\pi \times 4.00 \times 10^{-3} \mathrm{H}} \approx 7313.83 \ \mathrm{Hz}\)
03

Round the answer and write it in a suitable unit.

Round the frequency: \(f \approx 7.31 \times 10^3 \ \mathrm{Hz}\) Now, write the frequency in kilohertz (kHz): \(f \approx 7.31 \ \mathrm{kHz}\) So the frequency of the AC voltage source is approximately \(7.31 \ \mathrm{kHz}\).

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