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For an \(R C\) filter circuit, show that \(V_{\mathrm{R}}=V_{\mathrm{C}}=\mathcal{E}_{0} / \sqrt{2}\) at \(\omega=\omega_{c}\)

Short Answer

Expert verified
The voltages across the resistor and the capacitor in an RC circuit equate to \(\mathcal{E}_{0} / \sqrt{2}\) at the cut-off frequency \(\omega_{c}\).

Step by step solution

01

Understand the RC Circuit Relationship

In an RC circuit, the resistor voltage \(V_{R}\) is given by \(V_{R} = I R\), where \(I\) is the current and \(R\) is the resistance. Also, the capacitor voltage \(V_{C}\) can be represented by \(V_{C} = \mathcal{E}_{0} - V_{R}\), where \(\mathcal{E}_{0}\) is the source voltage.
02

Define the Cut-off Frequency

The cut-off frequency \(\omega_{c}\) in a RC circuit is calculated by \(\omega_{c} = 1 / R C\), where \(C\) is the capacitance of the capacitor. At this frequency, the outputs are reduced by a factor of \(1 / \sqrt{2}\), relative to the maximum output.
03

Recall the Impedance in an RC Circuit

The impedance \(Z\) of an RC circuit is given by \(Z = \sqrt{R^{2} + (1 / \omega C)^{2}}\). At the cut-off frequency \(\omega = \omega_{c}\), the impedance becomes \(Z = \sqrt{R^{2} + R^{2}} = R \sqrt{2}\). This gives the current \(I = \mathcal{E}_{0} / Z = \mathcal{E}_{0} / R \sqrt{2}\). Thus, both voltages \(V_{R} = V_{C} = I R = \mathcal{E}_{0} / \sqrt{2}\).
04

Proving the Voltage at Cut-off Frequency

Substituting the values derived in the previous steps: \(V_{R} = V_{C} = \mathcal{E}_{0} / \sqrt{2}\). This shows that at the cut-off frequency \(\omega_{c}\), both \(V_{R}\) and \(V_{C}\) are equals to \(\mathcal{E}_{0} / \sqrt{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Cut-off Frequency
In an RC circuit, the cut-off frequency, denoted as \( \omega_c \), is an essential parameter. It defines the point at which the output signal's power is reduced to half its maximum value. The cut-off frequency can be mathematically expressed as \( \omega_c = \frac{1}{RC} \), where \( R \) is the resistance and \( C \) is the capacitance.
At this frequency, the circuit's response changes significantly, making it a crucial point of analysis in filters and signal processing applications. For example, in a low-pass filter, frequencies below \( \omega_c \) are allowed to pass, while those above are attenuated. This behavior is critical in shaping signal frequencies according to the circuit's needs.
Exploring Impedance in an RC Circuit
Impedance is a measure of how much an RC circuit resists the flow of alternating current. It combines the effects of resistance and reactance into a single value. In our RC circuit, the impedance \( Z \) is calculated using the formula \( Z = \sqrt{R^2 + \left( \frac{1}{\omega C} \right)^2} \).
This formula shows impedance as a function of frequency \( \omega \), resistance \( R \), and capacitance \( C \). At the cut-off frequency \( \omega_c \), the reactance \( \frac{1}{\omega_c C} \) and resistance \( R \) combine such that the impedance reaches a special value \( R \sqrt{2} \). This value impacts how the current and voltage split between the resistor and capacitor, affecting their individual voltages.
Calculating Resistor Voltage
The resistor voltage in an RC circuit, denoted as \( V_R \), is directly related to both the current through the resistor and its resistance value. It's described by the formula \( V_R = IR \), where \( I \) is the current.
At the cut-off frequency, the impedance affects the current, giving \( I = \frac{\mathcal{E}_0}{R \sqrt{2}} \). The voltage across the resistor then becomes \( V_R = I R = \frac{\mathcal{E}_0}{\sqrt{2}} \). This relationship shows how at the critical frequency, the resistor's voltage is reduced, exemplifying the effect of filtering in the RC circuit.
Understanding Capacitor Voltage
Capacitor voltage, denoted as \( V_C \), highlights the potential difference across the capacitor in an RC circuit. It's calculated using the equation \( V_C = \mathcal{E}_0 - V_R \), where \( \mathcal{E}_0 \) is the source voltage.
At the cut-off frequency, when \( V_R = \frac{\mathcal{E}_0}{\sqrt{2}} \), the remaining voltage across the capacitor also equals \( \frac{\mathcal{E}_0}{\sqrt{2}} \). This equivalence at \( \omega_c \) illustrates how the energy is equally shared between the resistor and the capacitor, thereby confirming that both contribute to the circuit's overall behavior at the cut-off point. Understanding this balance is key to mastering how RC circuits manage voltages across their components.

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Most popular questions from this chapter

You're the operator of a \(15,000 \mathrm{V}\) rms, \(60 \mathrm{Hz}\) electrical substation. When you get to work one day, you see that the station is delivering 6.0 MW of power with a power factor of 0.90 a. What is the rms current leaving the station? b. How much series capacitance should you add to bring the power factor up to \(1.0 ?\) c. How much power will the station then be delivering?

Commercial electricity is generated and transmitted as threephase electricity. Instead of a single emf \(\mathcal{E}=\mathcal{E}_{0} \cos \omega t\), three separate wires carry currents for the emfs \(\mathcal{E}_{1}=\mathcal{E}_{0} \cos \omega t\) \(\mathcal{E}_{2}=\mathcal{E}_{0} \cos \left(\omega t+120^{\circ}\right),\) and \(\mathcal{E}_{3}=\mathcal{E}_{0} \cos \left(\omega t-120^{\circ}\right) .\) This is why the long-distance transmission lines you see in the countryside have three parallel wires, as do many distribution lines within a city. a. Draw a phasor diagram showing phasors for all three phases of a three-phase emf. b. Show that the sum of the three phases is zero, producing what is referred to as neutral. In single-phase electricity, provided by the familiar \(120 \mathrm{V} / 60 \mathrm{Hz}\) electric outlets in your home, one side of the outlet is neutral, as established at a nearby electrical substation. The other, called the hot side, is one of the three phases. (The round opening is connected to ground.) c. Show that the potential difference between any two of the phases has the rms value \(\sqrt{3} \mathcal{E}_{\text {rms }},\) where \(\mathcal{E}_{\text {rms }}\) is the familiar single-phase rms voltage. Evaluate this potential difference for \(\mathcal{E}_{\text {rms }}=120 \mathrm{V}\) Some high-power home appliances, especially electric clothes dryers and hot-water heaters, are designed to operate between two of the phases rather than between one phase and neutral. Heavy-duty industrial motors are designed to operate from all three phases, but full three phase power is rare in residential or office use.

a. For an \(R C\) circuit, find an expression for the angular frequency \(\omega_{\text {res }}\) at which \(V_{\mathrm{R}}=\frac{1}{2} \mathcal{E}_{0}\) b. What is \(V_{\mathrm{C}}\) at this frequency? c. What is \(\omega_{\text {res }}\) if the crossover frequency is \(6280 \mathrm{rad} / \mathrm{s} ?\)

\(\mathrm{A} 20 \mathrm{mH}\) inductor is connected across an AC generator that produces a peak voltage of \(10 \mathrm{V}\). What is the peak current through the inductor if the emf frequency is (a) \(100 \mathrm{Hz} ?\) (b) \(100 \mathrm{kHz} ?\)

A series RLC circuit attached to a \(120 \mathrm{V} / 60 \mathrm{Hz}\) power line draws 2.4 A of current with a power factor of \(0.87 .\) What is the value of the resistor?

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