Question

For the non-inverting amplifier shown in the figure, the input voltage is 1 V. The feedback network consists of 2 k$\Omega$ and 1 k$\Omega$ resistors as shown. 
If the switch is open, $V_o = x$. 
If the switch is closed, $V_o = ____ x$. 
 

Hint:

In a non-inverting amplifier, gain depends only on the ratio $\frac{R_f}{R_g}$. If a resistor in the feedback path is shorted, it effectively gets removed from the circuit, reducing the gain.

Answer 1

Correct Answer: 0.75

Step 1: Understanding the Question
This question analyzes a non-inverting operational amplifier circuit under two conditions: with a switch open and with it closed.
We need to calculate the output voltage in both cases and then express the second output voltage as a multiple of the first one ($x$).
Step 2: Key Formula or Approach
The circuit is a non-inverting amplifier.
The voltage gain ($A_v$) for this configuration is given by the formula:
\[ A_v = \frac{V_o}{V_{in}} = 1 + \frac{R_f}{R_g} \] where $R_f$ is the feedback resistance and $R_g$ is the resistor connected to the inverting terminal and ground.
Here, $V_{in} = 1\,\text{V}$ and $R_g = 1\,\text{k}\Omega$.
Step 3: Detailed Explanation
Case 1: Switch is Open
When the switch is open, the two resistors in the feedback path (2 k$\Omega$ and 1 k$\Omega$) are in series.
The total feedback resistance is:
\[ R_f = 2\,\text{k}\Omega + 1\,\text{k}\Omega = 3\,\text{k}\Omega \] The voltage gain in this case is:
\[ A_v = 1 + \frac{3\,\text{k}\Omega}{1\,\text{k}\Omega} = 1 + 3 = 4 \] The output voltage is:
\[ V_o = A_v \times V_{in} = 4 \times 1\,\text{V} = 4\,\text{V} \] The problem states that this output voltage is $x$. Therefore, $x = 4\,\text{V}$.
Case 2: Switch is Closed
When the switch is closed, it creates a short circuit across the 1 k$\Omega$ resistor in the feedback path.
This effectively removes the 1 k$\Omega$ resistor from the circuit, leaving only the 2 k$\Omega$ resistor as the feedback resistance.
\[ R_f = 2\,\text{k}\Omega \] The new voltage gain is:
\[ A_v = 1 + \frac{2\,\text{k}\Omega}{1\,\text{k}\Omega} = 1 + 2 = 3 \] The new output voltage is:
\[ V_o = A_v \times V_{in} = 3 \times 1\,\text{V} = 3\,\text{V} \] Step 4: Final Answer
We need to express the new output voltage (3 V) in terms of $x$ (where $x = 4\,\text{V}$).
\[ V_o = \frac{3}{4} \times 4\,\text{V} = \frac{3}{4} x \] Thus, when the switch is closed, $V_o = \frac{3}{4}x$.