/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 An ideal ac current source is ap... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 10

An ideal ac current source is applied to the input terminals of an amplifier, and the amplifier output voltage is \(2 \mathrm{~V} \mathrm{rms}\). Then, a \(2-\mathrm{k} \Omega\) resistance is placed in parallel with the current source and the amplifier input terminals, and the output voltage is \(1.5 \mathrm{~V}\) rms. Determine the input resistance of the amplifier.

Short Answer

Expert verified
The input resistance of the amplifier is \(6 k\Omega\).

Step by step solution

01

Understanding the Scenario

We have an ideal AC current source connected to an amplifier, which produces an output voltage of \(2 \mathrm{~V} \) RMS without any additional resistance. When a \(2 - k\Omega\) resistance is placed in parallel with the current source and amplifier input, the output voltage decreases to \(1.5 \mathrm{~V} \) RMS.
02

Analyzing the Effects of Added Resistance

The reduction in output voltage when the \(2 - k\Omega\) resistor is added is due to the input resistance of the amplifier forming a voltage divider with the added resistor. This affects the input voltage to the amplifier, thereby reducing the output voltage.
03

Setting Up the Voltage Divider Equation

Let \(R_{in}\) be the input resistance of the amplifier. The output voltage is affected by this resistor and the parallel \(2 - k\Omega\) resistor. We can set up the voltage divider equation: \( \frac{1.5}{2} = \frac{R_{in}}{R_{in} + 2000} \).
04

Solving for the Input Resistance

From the voltage divider equation: \( \frac{1.5}{2} = \frac{R_{in}}{R_{in} + 2000} \). First, solve for \(R_{in}:\)\( 1.5(R_{in} + 2000) = 2R_{in} \)This simplifies to: \( 1.5R_{in} + 3000 = 2R_{in} \)Rearranging gives: \( 2R_{in} - 1.5R_{in} = 3000 \)Which simplifies to: \( 0.5R_{in} = 3000 \)Finally solving for \(R_{in}\): \( R_{in} = \frac{3000}{0.5} = 6000 \Omega \).
05

Conclusion

The input resistance of the amplifier is \(6000 \Omega\) or \(6 k\Omega\). This matches the requirements of the scenario where the additional resistance affects the output voltage due to the voltage divider relationship.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Amplifier Input Resistance
When dealing with amplifiers, one important aspect to consider is the input resistance. The input resistance, often referred to as the impedance of the amplifier, is crucial because it determines how the amplifier interacts with the signal source. In simple terms, it can be thought of as the opposition the amplifier presents to the current flow into the amplifier input.
  • A higher input resistance is generally preferable as it results in less loading of the source which maintains the signal strength.
  • Low input resistance could lead to significant loss in the signal because more current gets diverted away from the amplifier.

In the context of the original exercise, the input resistance of the amplifier has been calculated to be approximately 6000 ohms or 6 k惟. This value explains the drop in output voltage when a 2 k惟 resistor is added in parallel, forming a voltage divider. The input resistance together with the external resistor determines how much of the input current actually reaches the amplifier, thus affecting the output voltage.
Voltage Divider
A voltage divider is a simple yet fundamental concept in electronics, especially when working with resistors and AC circuits. It involves two or more resistors in series, and it's used to produce a fraction of a voltage from a larger one. If we consider two resistors, the voltage across one of them is a fraction of the total voltage, determined by the ratio of the resistances.
  • The formula for a voltage divider is \( V_{out} = V_{in} \cdot \frac{R_2}{R_1 + R_2} \), where \( V_{out} \) is the output voltage, \( V_{in} \) is the input voltage, and \( R_1 \) and \( R_2 \) are the resistances.
  • This concept allows for control of voltage in a circuit without changing the source voltage.

In the exercise, placing a 2 k惟 resistance in parallel with the current source and amplifier input causes the system to behave as a voltage divider. The input resistance of 6 k惟 works with the 2 k惟 resistor to control the division of voltage, which affects how much voltage is applied to the amplifier input.
AC Current Source
An AC current source provides a constant sinusoidal current regardless of the load it is connected to. This is a cornerstone of AC circuits, and it's especially important in amplifiers as it can simulate real-world conditions for signal analysis.
  • The ideal AC current source adapts to changes in load without varying the current, which maintains consistent signal characteristics such as frequency and amplitude.
  • This allows the amplifier to process signals reliably, assuming the input resistance and other factors are appropriately managed.

In the original problem, an ideal AC current source drives the amplifier. Initially, without any additional resistors, the amplifier outputs a voltage of 2 V rms. However, the introduction of a parallel resistor affects the overall impedance seen by the current source, thus showing the practical impact of load changes in an AC circuit setup, something that must be considered carefully when designing and analyzing amplifier circuits.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An amplifier with \(R_{i}=2 \mathrm{k} \Omega, R_{o}=500 \Omega\), and \(R_{\text {moc }}=-10^{7} \Omega\) is operated with a 1-k \(\Omega\) load. A source having a Th茅venin resistance of \(1 \mathrm{k} \Omega\) and an open-circuit voltage of \(2 \cos (200 \pi t) \mathrm{mV}\) is connected to the input terminals. Determine the output voltage as a function of time and the power gain.

An amplifier has \(R_{i}=150 \Omega, R_{o}=2 \mathrm{k} \Omega\) and \(R_{m o c}=15 \mathrm{k} \Omega\). Determine the values (including units) of \(A_{\mathrm{voc}}, G_{m \mathrm{sc}}\), and \(A_{\text {isc }}\) for this amplifier.

A certain amplifier has a differential gain of 1000 . If the two input terminals are tied together and a \(5-\mathrm{mV}-\mathrm{rms}\) input signal is applied, the output signal is \(30 \mathrm{mV} \mathrm{rms}\). Find the CMRR for this amplifier.

A certain amplifier operating with a \(150-\Omega\) load has a voltage gain of 75 and a power gain of 8000 . Determine the current gain and input resistance of the amplifier.

The input signal to an amplifier is \(v_{i}(t)=\) \(0.01 \cos (2000 \pi t)+0.02 \cos (4000 \pi t)\). The gain of the amplifier as a function of frequency is given by $$ A=\frac{100}{1+j(f / 1000)} $$ Find an expression for the output signal of the amplifier as a function of time.
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Measurements

Read Explanation

Space Physics

Read Explanation

Physics of Motion

Read Explanation

Radiation

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.