Question

(a) What is the reactance of a \(10.0-\mathrm{mH}\) inductor at the frequency \(f=250.0 \mathrm{Hz} ?\) (b) What is the impedance of a series combination of the \(10.0-\mathrm{mH}\) inductor and a \(10.0-\Omega\) resistor at $250.0 \mathrm{Hz} ?$ (c) What is the maximum current through the same circuit when the ac voltage source has a peak value of \(1.00 \mathrm{V} ?\) (d) By what angle does the current lag the voltage in the circuit?

Short answer

Question: Calculate the reactance of the inductor, the impedance of the series combination of an inductor and a resistor, the maximum current in the circuit, and the phase angle between the current and voltage when the inductor has a value of 10.0 mH, the resistor has a value of 10.0 Ω, the frequency is 250.0 Hz, and the peak voltage is 1.00 V. Answer: The reactance of the inductor at 250.0 Hz is 15.7 Ω, the impedance of the series combination is approximately 18.8 Ω, the maximum current in the circuit is approximately 0.053 A, and the phase angle between the current and voltage is approximately 57.6°.

Step-by-step solution

Calculate the inductive reactance

To find the inductive reactance (\(X_L\)), we can use the formula: \(X_L = 2\pi fL\) where \(f = 250.0 \mathrm{Hz}\) and \(L = 10.0 \mathrm{mH}\). Plug in the values and solve for \(X_L\). \(X_L = 2\pi (250.0 \mathrm{Hz})(10.0 \times 10^{-3} \mathrm{H})\) \(X_L = 15.7 \Omega\) The reactance of the inductor at \(250.0 \mathrm{Hz}\) is \(15.7 \Omega\).

Calculate the impedance

To find the impedance (Z) of the series combination of the inductor and the resistor, we can use the formula: \(Z = \sqrt{R^2 + X_L^2}\) where \(R = 10.0 \Omega\) and \(X_L = 15.7 \Omega\). Plug in the values and solve for Z: \(Z = \sqrt{(10.0 \Omega)^2 + (15.7 \Omega)^2}\) \(Z \approx 18.8 \Omega\) The impedance of the series combination at \(250.0 \mathrm{Hz}\) is approximately \(18.8 \Omega\).

Calculate the maximum current

To find the maximum current (I) in the circuit, we can use the formula: \(I = \frac{V}{Z}\) where \(V = 1.00 \mathrm{V}\) and \(Z \approx 18.8 \Omega\). Plug in the values and solve for I: \(I \approx \frac{1.00 \mathrm{V}}{18.8 \Omega}\) \(I \approx 0.053 \mathrm{A}\) The maximum current in the circuit is approximately \(0.053 \mathrm{A}\).

Calculate the phase angle

To find the phase angle (\(\theta\)) between the current and voltage in the circuit, we can use the formula: \(\theta = \tan^{-1}(\frac{X_L}{R})\) where \(X_L = 15.7 \Omega\) and \(R = 10.0 \Omega\). Plug in the values and solve for \(\theta\): \(\theta = \tan^{-1}(\frac{15.7 \Omega}{10.0 \Omega})\) \(\theta \approx 57.6^{\circ}\) The phase angle between the current and voltage is approximately \(57.6^{\circ}\). The current lags the voltage by this angle in the circuit.