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: Identify the configuration.
The circuit is a non-inverting amplifier.
For a non-inverting amplifier, the gain is given by:
\[ A_v = 1 + \frac{R_f}{R_g} \] where $R_f$ is the feedback resistance (from output to inverting terminal) and $R_g$ is the resistance from inverting terminal to ground.
Here, $R_g = 1\,\text{k}\Omega$.
Step 2: Case 1 — Switch Open.
When the switch is open, the 2 k$\Omega$ and 1 k$\Omega$ resistors are in series.
Therefore, total feedback resistance:
\[ R_f = 2\,\text{k}\Omega + 1\,\text{k}\Omega = 3\,\text{k}\Omega. \] Now the gain becomes:
\[ A_v = 1 + \frac{3}{1} = 4. \] Since input voltage is 1 V:
\[ V_o = 4 \times 1 = 4\,\text{V}. \] So,
\[ x = 4\,\text{V}. \]
Step 3: Case 2 — Switch Closed.
When the switch is closed, the 1 k$\Omega$ resistor gets short-circuited.
Thus only 2 k$\Omega$ remains in the feedback path.
So,
\[ R_f = 2\,\text{k}\Omega. \] Now the gain becomes:
\[ A_v = 1 + \frac{2}{1} = 3. \] Thus output voltage becomes:
\[ V_o = 3 \times 1 = 3\,\text{V}. \]
Step 4: Express in terms of $x$.
We found:
\[ x = 4\,\text{V}. \] When switch is closed, output is 3 V.
Therefore,
\[ V_o = \frac{3}{4} x. \]