For the difference amplifier in Fig. 6_4, if V₁ = 3 sin10t (V), V₂= 6 sin10t (V), and R= 20 Ohm, what is Vout? V₁ V₂ 3 sin10t (V) -3 sin10t (V) -9 sin10t (V) 9 sin 10t (V) R R R Fig. 6_4 R ww Vout RL
For the difference amplifier in Fig. 6_4, if V₁ = 3 sin10t (V), V₂= 6 sin10t (V), and R= 20 Ohm, what is Vout? V₁ V₂ 3 sin10t (V) -3 sin10t (V) -9 sin10t (V) 9 sin 10t (V) R R R Fig. 6_4 R ww Vout RL
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Question
![**Transcription for Educational Website**
---
**Problem Statement:**
For the difference amplifier in Fig. 6_4, if \( V_1 = 3 \sin(10t) \, (V) \), \( V_2 = 6 \sin(10t) \, (V) \), and \( R = 20 \, \Omega \), what is \( V_{\text{out}} \)?
**Diagram Description:**
- The diagram shows a difference amplifier circuit configuration.
- It includes two voltage sources \( V_1 \) and \( V_2 \).
- Both inputs \( V_1 \) and \( V_2 \), are connected through resistors labeled \( R \) to the inverting and non-inverting inputs of an operational amplifier (op-amp).
- The op-amp has feedback resistance \( R \) and another \( R \) connected to ground from the non-inverting input.
- The output voltage is labeled \( V_{\text{out}} \), with \( R_L \) connecting it to ground.
**Answer Choices:**
- \( \circ \) \( 3 \sin(10t) \, (V) \)
- \( \circ \) \(-3 \sin(10t) \, (V) \)
- \( \circ \) \(-9 \sin(10t) \, (V) \)
- \( \circ \) \( 9 \sin(10t) \, (V) \)
---
**Solution Explanation:**
To determine \( V_{\text{out}} \) for a difference amplifier, use the formula:
\[
V_{\text{out}} = \left( \frac{R_f}{R_1} \right) (V_2 - V_1)
\]
In this case, all resistors are equal (\( R = R_f = R_1 \)), so:
\[
V_{\text{out}} = V_2 - V_1
\]
Substituting the given values:
\[
V_{\text{out}} = 6 \sin(10t) - 3 \sin(10t) = 3 \sin(10t)
\]
This corresponds to the answer choice:
\( \circ \) \( 3 \sin(10t) \, (V) \)
Thus, the correct answer is \( 3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F849b5f85-1e18-473b-8772-ae9929e740d6%2F87f789b1-db9a-4888-bc9e-9749f8732274%2F0a6rf26_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
---
**Problem Statement:**
For the difference amplifier in Fig. 6_4, if \( V_1 = 3 \sin(10t) \, (V) \), \( V_2 = 6 \sin(10t) \, (V) \), and \( R = 20 \, \Omega \), what is \( V_{\text{out}} \)?
**Diagram Description:**
- The diagram shows a difference amplifier circuit configuration.
- It includes two voltage sources \( V_1 \) and \( V_2 \).
- Both inputs \( V_1 \) and \( V_2 \), are connected through resistors labeled \( R \) to the inverting and non-inverting inputs of an operational amplifier (op-amp).
- The op-amp has feedback resistance \( R \) and another \( R \) connected to ground from the non-inverting input.
- The output voltage is labeled \( V_{\text{out}} \), with \( R_L \) connecting it to ground.
**Answer Choices:**
- \( \circ \) \( 3 \sin(10t) \, (V) \)
- \( \circ \) \(-3 \sin(10t) \, (V) \)
- \( \circ \) \(-9 \sin(10t) \, (V) \)
- \( \circ \) \( 9 \sin(10t) \, (V) \)
---
**Solution Explanation:**
To determine \( V_{\text{out}} \) for a difference amplifier, use the formula:
\[
V_{\text{out}} = \left( \frac{R_f}{R_1} \right) (V_2 - V_1)
\]
In this case, all resistors are equal (\( R = R_f = R_1 \)), so:
\[
V_{\text{out}} = V_2 - V_1
\]
Substituting the given values:
\[
V_{\text{out}} = 6 \sin(10t) - 3 \sin(10t) = 3 \sin(10t)
\]
This corresponds to the answer choice:
\( \circ \) \( 3 \sin(10t) \, (V) \)
Thus, the correct answer is \( 3
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