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Chapter 14: Problem 2097

For high frequency, a capacitor offers (a) More reactance (b) Less reactance (c) Zero reactance (d) Infinite reactance

Short Answer

Expert verified
(b) Less reactance

Step by step solution

01

1. Understanding Capacitive Reactance

Capacitive reactance refers to the capacitor's resistance to alternating current (AC) signals. This resistance changes based on the frequency of the input signal. To understand how a capacitor behaves with high-frequency signals, we need to consider the formula for capacitive reactance, which is given by: \(X_C = \frac{1}{2\pi fC}\) Here, \(X_C\) is the capacitive reactance, \(f\) is the frequency of the signal, and \(C\) is the capacitor's capacitance.
02

2. Analyzing the Capacitive Reactance Formula for High Frequencies

Let's consider the capacitive reactance formula and analyze how \(X_C\) changes based on the frequency \(f\). Since \(X_C\) is inversely proportional to the frequency, we can deduce that as the frequency of the input signal increases, the capacitive reactance decreases, and vice versa. Therefore, for high-frequency signals, a capacitor would offer a lower reactance. Now that we have analyzed the behavior of a capacitor for high-frequency signals, we can choose the correct option from the given alternatives.
03

3. Final Selection

Based on our analysis, we can select the correct option: (b) Less reactance A capacitor offers less reactance for high-frequency input signals.

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Most popular questions from this chapter

An inductor of inductance \(\mathrm{L}\) and resistor of resistance \(\mathrm{R}\) are joined in series and connected by a source of frequency 0 power dissipated in the circuit is, (a) $\left[\left\\{\mathrm{R}^{2}+\mathrm{c}^{2} \mathrm{~L}^{2}\right\\} / \mathrm{V}\right]$ (b) $\left[\left\\{\mathrm{V}^{2} \mathrm{R}\right\\} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (c) $\left[\mathrm{V} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (d) $\left[\sqrt{ \left.\left\\{R^{2}+\omega^{2} L^{2}\right\\} / V^{2}\right]}\right.$

A magnetic field \(2 \times 10^{-2} \mathrm{~T}\) acts at right angles to a coil of area \(200 \mathrm{~cm}^{2}\) with 25 turns. The average emf induced in the coil is \(0.1 \mathrm{v}\) when it removes from the field in time t. The value of \(\mathrm{t}\) is (a) \(0.1 \mathrm{sec}\) (b) \(1 \mathrm{sec}\) (c) \(0.01 \mathrm{sec}\) (d) \(20 \mathrm{sec}\)

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