Question

In the given circuit, with an ideal op-amp for what value of \(\frac{R_1}{R_2}\) the output of the amplifier V_out = V₂ - V₁ ?

• A. 1
• B. \(\frac{1}{2}\)
• C. 2
• D. \(\frac{3}{2}\)

Answer 1

The Correct Option is A

To find the value of \(\frac{R_1}{R_2}\) for which the output of the amplifier \(V_{\text{out}} = V_2 - V_1\), let's analyze the given circuit with the ideal operational amplifier (op-amp).

In this circuit, we have: 

• An inverting input connected through resistor \(R\) to \(V_1\).
• A non-inverting input connected to \(V_2\) via resistors \(R_1\) and \(R_2\).

The ideal op-amp has the property that the voltage difference between the inverting and non-inverting inputs is zero when feedback is present (virtual short circuit concept).

For a differential amplifier configuration, the output voltage \(V_{\text{out}}\) is given by:

\(V_{\text{out}} = \left(1 + \frac{R}{R}\right)(V_2 - V_1)\)

Since the resistors used are identical in input and feedback paths, the gain is only dependent on the resistors \(R_1\) and \(R_2\). The configuration given in the image works when it forms a differential amplifier with unity gain, i.e.:

\(V_{\text{out}} = V_2 - V_1\)

This happens when:

\(\frac{R_1}{R_2} = 1\)

The output directly subtracts \(V_1\) from \(V_2\) when \(R_1 = R_2\).

Conclusion:

Therefore, the correct answer is:

1