Question

In a series RLC circuit, \(V=(12.0 \mathrm{~V})(\sin \omega t), R=10.0 \Omega, L=2.00 \mathrm{H}\) and \(C=10.0 \mu \mathrm{F}\). At resonance, determine the amplitude of the voltage across the inductor. Is the result reasonable, considering that the voltage supplied to the entire circuit has an amplitude of \(12.0 \mathrm{~V} ?\)

Short answer

Question: Calculate the amplitude of the voltage across the inductor at resonance in an RLC circuit with given resistance R = 10.0 Ω, inductance L = 2.00 H, capacitance C = 10.0 µF, and a sinusoidal input voltage amplitude of 12.0 V. Solution: Step 1: Calculate the angular frequency at resonance: ω_0 = (1 / √((2.00 H)(10.0×10^-6 F))) = 1125 rad/s Step 2: Calculate the impedance of the RLC circuit at resonance: Z = R = 10.0 Ω Step 3: Calculate the current through the circuit at resonance: I = (12.0 V) / (10.0 Ω) = 1.20 A Step 4: Calculate the voltage across the inductor at resonance: |V_L| = I × ω_0L = (1.20 A) × (1125 rad/s)(2.00 H) = 2700 V Step 5: Check if the result is reasonable: Since the calculated voltage amplitude across the inductor (2700 V) is much greater than the given amplitude of the input voltage (12.0 V), the result may seem unreasonable. However, this is due to the high inductance and capacitance values in this RLC circuit, which leads to a high resonance voltage across the inductor at resonance frequency. Thus, the calculated result of 2700 V is correct.

Step-by-step solution

Calculate the angular frequency at resonance

At resonance, the impedance of the circuit is just the resistive component, so the angular frequency of the source is the resonance frequency: \(\omega_0 = \dfrac{1}{\sqrt{LC}}\) Plug in the given values of L and C: \(\omega_0 = \dfrac{1}{\sqrt{(2.00 H)(10.0\times10^{-6}F)}}\) and solve for \(\omega_0\).

Calculate the impedance of the RLC circuit at resonance

Given R, L and C, the impedance at resonance can be calculated using formula: \(Z = R\) Plug in the given value of R: \(Z = 10.0\Omega\)

Calculate the current through the circuit at resonance

Using Ohm's Law, \(V = IZ\), we can find the current by rearranging the formula: \(I = \dfrac{V}{Z}\) Plug in the given value of the sinusoidal input amplitude V and the impedance Z found in step 2: \(I = \dfrac{(12.0V)}{(10.0\Omega)}\)

Calculate the voltage across the inductor at resonance

Now, we can calculate voltage across the inductor using Ohm's Law at resonance: \(V_L = I \times Z_L = I \times (j\omega_0L)\) We only need the amplitude for the voltage, so we’ll consider only the magnitude of the voltage: \(|V_L| = I \times \omega_0L\) Plug in the values of I, L, and \(\omega_0\) found in previous steps to find the magnitude of the voltage across the inductor at resonance.

Check if the result is reasonable

Compare the calculated voltage amplitude across the inductor with the given amplitude of the input voltage (12.0 V). If the calculated amplitude is less than or equal to the input voltage amplitude, then the result is reasonable.

Key concepts

Resonance Frequency in Series RLC Circuits

In the realm of electrical engineering, resonance frequency marks a pivotal concept, especially within the confines of a series RLC circuit. This is the frequency at which the reactive effects of the circuit's inductor (L) and capacitor (C) perfectly balance out, resulting in the overall impedance being purely resistive.

Mathematically, the resonance frequency in an RLC circuit is defined by the formula:
\begin{align*} f_0 = \dfrac{1}{2\pi\sqrt{LC}} \end{align*}
where:\

• \( f_0 \) is the resonance frequency in hertz,
• \( L \) is the inductance in henrys,
• \( C \) is the capacitance in farads.

At this frequency, inductive and capacitive reactances are equal in magnitude but opposite in sign, canceling each other out. As a result, the circuit behaves as if it only contains a resistance, making it highly efficient for the AC power source.

Angular Frequency and Its Importance

Diving deeper into the nuances of alternating currents (AC), angular frequency emerges as a critical parameter. It represents the oscillation rate of the sinusoidal waveform of the current or voltage in radians per second and is related to the resonance frequency by a simple multiplier of \( 2\pi \), denoted as:\
\begin{align*} \omega = 2\pi f \end{align*}
For our RLC circuit, the angular frequency at resonance is determined through the formula:\
\begin{align*} \omega_0 = \dfrac{1}{\sqrt{LC}} \end{align*}
which is derived from the resonance frequency formula by incorporating the \( 2\pi \) factor. The angular frequency is critical because it allows us to understand and calculate the behavior of the circuit elements, especially the inductors and capacitors, in response to the changing AC supply.

Impedance in a Resonant Series RLC Circuit

Understanding impedance is key to grasping the behavior of RLC circuits. Impedance, denoted by \( Z \), is the total opposition a circuit presents to alternating current, comprising both resistance (R) and reactance (X). Its unit is the ohm (\( \Omega \)).

In a series RLC circuit at resonance, the inductive reactance \( (X_L) \) and capacitive reactance \( (X_C) \) cancel each other out, leaving us with:\
\begin{align*} Z = R \end{align*}
This simplification results in maximum current flow for a given voltage. Thus, at resonance, the circuit exhibits its lowest impedance and behaves as if it were purely resistive.

Ohm's Law and Its Application

At the heart of circuit analysis lies Ohm's Law, a foundational principle allowing us to analyze electrical circuits—both in components and as a whole. Expressed simply, Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them: \[ I = \frac{V}{R} \]

Applying this law within the context of our series RLC circuit at resonance, we're able to calculate the current flow once we establish that the impedance is purely resistive. Utilizing the aforementioned relationship of \( V = IZ \) at resonance simplifies the calculus since \( Z = R \). This allows us to easily deduce the current, using the voltage supplied to the circuit and the resistance of the circuit, and ultimately compute the voltage across any component, such as the inductor, by multiplying the current by the component's impedance.