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

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**Transcription for Educational Website**

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**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:**

- \( \circ \) \( 3 \sin(10t) \, (V) \)
- \( \circ \) \(-3 \sin(10t) \, (V) \)
- \( \circ \) \(-9 \sin(10t) \, (V) \)
- \( \circ \) \( 9 \sin(10t) \, (V) \)

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**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
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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