Question

A circuit contains a \(100 .-\Omega\) resistor, a 0.0500 - H inductor, a 0.400 - \(\mu\) F capacitor, and a source of time-varying emf connected in series. The time-varying emf corresponds to \(V_{\mathrm{rms}}=50.0 \mathrm{~V}\) at a frequency of \(2000 . \mathrm{Hz}\) a) Determine the current in the circuit. b) Determine the voltage drop across each component of the circuit. c) How much power is drawn from the source of emf?

Short answer

**Answer**: To find the rms current, voltage drop across each component, and power drawn from the time-varying emf source in the given RLC circuit, follow these steps: 1. Calculate the angular frequency, \(\omega\), using the given frequency. 2. Find the impedance of the inductor, \(Z_L\), and capacitor, \(Z_C\), using their respective formulas. 3. Determine the total circuit impedance, \(Z\), by adding the impedance of the resistor, inductor, and capacitor. 4. Calculate the rms current, \(I_{\text{rms}}\), using the rms voltage and the magnitude of the total impedance, \(|Z|\). 5. Determine the voltage drop across each component using Ohm's Law. 6. Calculate the average power drawn from the emf source using the rms current and resistance. By following these steps, you will be able to find the rms current, voltage drop across each component, and power drawn from the time-varying emf source in the RLC circuit.

Step-by-step solution

Calculate Angular Frequency

Given frequency is \(2000\:Hz\). Angular frequency can be calculated as: \(\omega = 2\pi f\), where \(f\) is the frequency. $$\omega = 2\pi (2000)=4000\pi\:rad/s$$

Calculate the Impedances of Inductor and Capacitor

For the inductor, impedance is given by \(Z_L=j\omega L\), where \(L\) is the inductance and \(j\) is the imaginary unit. $$Z_L=j(4000\pi)(0.0500)=200\pi j\: \Omega$$ For the capacitor, impedance is given by \(Z_C=\frac{1}{j\omega C}\), where \(C\) is the capacitance. $$ Z_C = \frac{1}{j(4000\pi)(0.400\times10^{-6})} = \frac{1}{1.6\pi j} = -j\frac{5\pi}{32} \: \Omega$$

Determine Circuit Impedance

The total impedance in the circuit can be found using the formula: \(Z = Z_R+Z_L+Z_C\) $$Z = 100 + 200\pi j - j\frac{5\pi}{32}=100+(200\pi-\frac{5\pi}{32})j \ \Omega$$

Calculate RMS Current

We can determine the rms current using the equation: \(I_{\text{rms}}=\frac{V_{\text{rms}}}{|Z|}\), where \(|Z|\) is the magnitude of total impedance \(Z\). $$|Z| = \sqrt{(100)^2 + (200\pi - \frac{5\pi}{32})^2}$$ $$I_{\text{rms}} = \frac{50}{|Z|}$$

Voltage Drops

Voltage drops across the resistor, inductor, and capacitor can be calculated using Ohm's Law: \(V=IZ\). $$V_R = I_\text{rms} Z_R$$ $$V_L = I_\text{rms} Z_L$$ $$V_C = I_\text{rms} Z_C$$

Calculate Power

The average power drawn from the emf source can be determined using the equation: \(P_\text{avg}=I^{2}_{\text{rms}} R\) $$P_\text{avg} = (I_{\text{rms}})^2(100)$$ In conclusion, we have determined the current in the circuit, voltage drop for each component, and the power drawn from the emf source.

Key concepts

AC Circuit Impedance

In an alternating current (AC) circuit, impedance plays the role of resistance in limiting the flow of current, but it also accounts for the fact that current and voltage are not always in phase. Impedance combines resistance (R), inductive reactance (X_L), and capacitive reactance (X_C) into a single complex quantity that describes the opposition that a circuit presents to the flow of alternating current. The inductive and capacitive reactances are frequency-dependent. In our problem, the impedance across the inductor and capacitor were calculated individually and then combined with the resistor's impedance to find the total impedance, which is paramount to determining the current and voltage distributions in the circuit.

Understanding the total impedance is essential for analyzing RLC circuits, making it possible to predict how the circuit will respond to different frequencies of AC power.

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering and physics that states the current through a conductor between two points is directly proportional to the voltage across the two points. Expressed with the formula V = IR, where V is voltage, I is current, and R is resistance. For AC circuits, Ohm's Law applies to impedance instead of just resistance. It's crucial to remember that Ohm's Law works with the RMS (root mean square) values of AC circuits to accommodate the varying nature of the current and voltage. Therefore, when we calculated the voltage drops across each component in our exercise, we applied Ohm's Law using the impedance specific to each component and the RMS value of the current.

Angular Frequency

Angular frequency, often denoted by the Greek letter omega (\r \( \r \omega \)), is a measure of how quickly an AC circuit oscillates and is related to the traditional frequency (f)—the number of cycles per second—by the equation \( \r \omega = 2\r \pi \f \). It is measured in radians per second and gives invaluable information when analyzing AC circuits, particularly resonant circuits which consist of inductors and capacitors. In our problem, angular frequency helped us determine the impedances of the inductor and capacitor, reflecting how these components affect the circuit at a given frequency. High angular frequency can lead to significant changes in the behavior of these components.

RMS Current

Root Mean Square (RMS) current is a statistical measure of the magnitude of a varying current. It represents the equivalent direct current (DC) value that delivers the same power to a load as the AC current. The RMS value is crucial in AC circuit analysis, as it allows for the calculation of power consumption and voltage drops for components that operate with varying current and voltage. The RMS current in our circuit was calculated from the total impedance and the RMS voltage of the source. This technique is commonly used in household and industrial applications where AC is the form of electricity.

Circuit Power Consumption

Power consumption in an AC circuit is not as straightforward as in a DC circuit due to the alternating nature of the voltage and current. The power consumed by an AC circuit is given by \( P_{avg} = I_{rms}^2 R \), which considers only the real part of the impedance (the resistance in this case). Our solved problem needed the power drawn from an AC source, which requires calculating the RMS current first and then applying it to the formula along with the resistor's value. This gives us the average power consumed by the circuit, which is crucial for designing and managing electrical systems, ensuring that components are not overloaded and that energy efficiency is optimized.