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(II) A resistor \(R\) is in parallel with a capacitor \(C ,\) and this parallel combination is in series with a resistor \(R ^ { \prime }\) . If connected to an ac voltage source of frequency \(\omega ,\) what is the equivalent impedance of this circuit at the two extremes in frequency \(( a ) \omega = 0 ,\) and \(( b ) \omega = \infty ?\)

Short Answer

Expert verified
For \( \omega = 0 \), impedance is \( R + R' \); for \( \omega = \infty \), impedance is \( R' \).

Step by step solution

01

Understanding Circuit Configuration

In this problem, we have a parallel connection of a resistor \( R \) and a capacitor \( C \), which is further in series with another resistor \( R' \). The task involves finding the equivalent impedance of this arrangement when subjected to an AC voltage source at two extreme frequencies: \( \omega = 0 \) and \( \omega = \infty \).
02

Calculate Impedance at \( \omega = 0 \)

At \( \omega = 0 \), the frequency is zero, which means it behaves like a DC circuit. The capacitor acts as an open circuit at DC. Thus, the impedance of the capacitor is infinite and effectively only the resistor \( R \) contributes to the parallel section. The equivalent impedance here is \( R \) in parallel with an open circuit, simplifying to \( R \), and then in series with \( R' \). Therefore, the total equivalent impedance is \( R + R' \).
03

Calculate Impedance at \( \omega = \infty \)

At \( \omega = \infty \), the frequency is very high. Here, the capacitor behaves like a short circuit. Consequently, the parallel combination of \( R \) and the capacitor tends towards a short circuit, meaning the impedance of this branch is zero. Thus, the equivalent impedance seen by the source is just \( R' \).
04

Summarize Results

For \( \omega = 0 \), the equivalent impedance is \( R + R' \). For \( \omega = \infty \), the equivalent impedance is \( R' \).

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Key Concepts

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

Resistor in Parallel with Capacitor
When you have a resistor in parallel with a capacitor, it means that these two components share both their connection points. In this setup, electrical current can flow through either the resistor or the capacitor. How each component behaves depends heavily on the frequency of the input signal.

At low frequencies, or when analyzing with direct current (DC), the capacitor acts more like an open circuit since it ideally blocks DC after it is fully charged. Therefore, at these frequencies, the current flows primarily through the resistor. However, at higher frequencies, especially in cases of alternating current (AC), the capacitor starts to act as a short circuit, allowing the AC to easily pass through it.

This behavior is crucial when understanding AC circuit analysis, as it directly influences the equivalent impedance.
AC Circuit Analysis
AC circuit analysis is all about understanding how alternating current flows through circuits. It's a bit different from analyzing direct current circuits because the current and voltages change direction and amplitude with time.

In AC circuits, elements such as resistors, capacitors, and inductors behave differently compared to DC circuits. For instance, in an AC circuit, the frequency affects the impedance, which is the combination of resistance, inductance, and capacitance. The impedance varies because capacitors and inductors store energy temporarily and release it back to the circuit, causing phase shifts between current and voltage.

To handle these complexities, concepts like phasors and impedance are used. Phasors allow us to visualize AC voltages and currents as rotating vectors. Impedance combines the effects of resistance and reactance (which includes both capacitive and inductive reactance) into one simple value, using complex numbers usually in polar or rectangular form.
Impedance at Extreme Frequencies
The impedance of a circuit describes how it resists the flow of alternating current, and this resistance changes with frequency, especially in circuits with reactive components like capacitors.

At extreme frequencies, such as when frequency approaches zero or infinity, a circuit's behavior simplifies. At a frequency approaching zero (DC conditions), a capacitor's impedance is theoretically infinite, acting like it’s not even there, while resistors maintain their standard resistance. This results in the impedance being mainly about the resistors in the circuit. For a parallel resistor-capacitor setup at zero frequency, this makes the capacitor invisible, leaving the resistor as the main component of impedance along with any series resistors.

Conversely, at a very high frequency, the capacitor behaves like a short circuit, essentially erasing its own presence in terms of impedance. This changes the focus to any resistors left in series. Understanding these boundary conditions helps simplify complex AC circuit analysis into manageable scenarios.

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Most popular questions from this chapter

A 2200 -pF capacitor is charged to 120\(\mathrm { V }\) and then quickly connected to an inductor. The frequency of oscillation is observed to be 17\(\mathrm { kHz }\) . Determine \(( a )\) the inductance, (b) the peak value of the current, and (c) the maximum energy stored in the magnetic field of the inductor.

A pair of straight parallel thin wires, such as a lamp cord, each of radius \(r,\) are a distance \(\ell\) apart and carry current to a circuit some distance away. Ignoring the field within each wire, show that the inductance per unit length is \(\left(\mu_{0} / \pi\right) \ln [(\ell-r) / r]\).

(II) A capacitor is placed in parallel with some device, B as in Fig. \(18 b ,\) to filter out stray high frequency signals, but to allow ordinary \(60 - H z\) ac to pass through with little loss. Suppose that circuit \(B\) in Fig. 18\(b\) is a resistance \(R = 490 \Omega\) connected to ground, and that \(C = 0.35 \mu \mathrm { F }\) . What percent of the incoming current will pass through \(C\) rather than \(R\) if it is \(( a )\) it is 60\(\mathrm { Hz }\) ; (b) \(60,000 \mathrm { Hz }\) ?

(I) \(\mathrm { A } 3800\) -pF capacitor is connected in series to a \(26.0 - \mu \mathrm { H }\) coll of resistance 2.00\(\Omega .\) What is the resonant frequency of this circuit?

A \(75-\Omega\) resistor and a \(6.8-\mu \mathrm{F}\) capacitor are connected in series to an ac source. Calculate the impedance of the circuit if the source frequency is \((a) 60 \mathrm{~Hz} ;(b) 6.0 \mathrm{MHz}\)

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