Question

In a certain RLC circuit, a \(20.0-\Omega\) resistor, a \(10.0-\mathrm{mH}\) inductor, and a \(5.00-\mu F\) capacitor are connected in series with an AC power source for which \(V_{\mathrm{rms}}=10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz}\). Calculate a) the amplitude of the current, b) the phase between the current and the voltage, and c) the maximum voltage across each component.

Short answer

Question: In an RLC circuit with a resistor of \(20\, \Omega\), an inductor of \(10\, mH\), and a capacitor of \(5\, \mu F\), connected in series with an AC power source of \(10\, Vrms\) at \(100\, Hz\), find the amplitude of the current, the phase between the current and voltage, and the maximum voltage across each component. Answer: The amplitude of the current is \(\frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}} \, \mathrm{A}\), the phase between the current and voltage is \(\tan^{-1}(-4\pi) \, \mathrm{rad}\), and the maximum voltage across each component is: - \(V_R = \frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}}(20) \, \mathrm{V}\) - \(V_L = \frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}}(20\pi) \, \mathrm{V}\) - \(V_C = \frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}}(100\pi) \, \mathrm{V}\)

Step-by-step solution

Compute the angular frequency

The first step is to calculate the angular frequency \((\omega)\) of the AC power source, which is given by: $$\omega = 2\pi f$$ where \(f\) is the frequency. In this case, \(f = 100\) Hz. Therefore, the angular frequency is: $$\omega = 2 \pi (100) = 200 \pi \, \mathrm{rad/s}$$

Calculate the impedance of the circuit

Next, we will find the impedance \((Z)\) of the circuit, which is given by the formula: $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ where \(R\) is the resistance, \(X_L = \omega L\) is the inductive reactance, \(X_C = \frac{1}{\omega C}\) is the capacitive reactance, \(L\) is the inductance, and \(C\) is the capacitance. 1. Compute \(X_L\): $$X_L = \omega L = (200\pi)(10\times10^{-3}) = 20\pi \, \Omega$$ 2. Compute \(X_C\): $$X_C = \frac{1}{\omega C} = \frac{1}{(200\pi)(5\times10^{-6})} = 100\pi \, \Omega$$ 3. Compute \(Z\): $$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(20)^2 + (20\pi - 100\pi)^2} = \sqrt{400 + 6400\pi^2} \, \Omega$$

Calculate the amplitude of the current

Now, we can calculate the amplitude of the current \((I_\mathrm{max})\) using Ohm's law: $$I_\mathrm{max} = \frac{V_\mathrm{max}}{Z}$$ where \(V_\mathrm{max} = \sqrt{2} V_{\mathrm{rms}}\) is the maximum voltage. 1. Compute \(V_\mathrm{max}\): $$V_\mathrm{max} = \sqrt{2} (10) = 10\sqrt{2} \, \mathrm{V}$$ 2. Compute \(I_\mathrm{max}\): $$I_\mathrm{max} = \frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}} \, \mathrm{A}$$

Find the phase between the current and voltage

The phase between the current and voltage \((\phi)\) can be calculated using the formula: $$\phi = \tan^{-1} \left(\frac{X_L - X_C}{R}\right)$$ Compute \(\phi\): $$\phi = \tan^{-1} \left(\frac{20\pi - 100\pi}{20}\right) = \tan^{-1}(-4\pi) \, \mathrm{rad}$$

Calculate the maximum voltage across each component

The maximum voltage across each component can be calculated using the following formulas: 1. Resistor: \(V_R = I_\mathrm{max} R\) 2. Inductor: \(V_L = I_\mathrm{max} X_L\) 3. Capacitor: \(V_C = I_\mathrm{max} X_C\) Compute \(V_R\), \(V_L\), and \(V_C\): $$V_R = \frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}}(20) \, \mathrm{V}$$ $$V_L = \frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}}(20\pi) \, \mathrm{V}$$ $$V_C = \frac{10\sqrt{2}}{\sqrt{400 + 6400\pi^2}}(100\pi) \, \mathrm{V}$$