In the circuit shown in the figure, both OPAMPs are ideal. The output for the circuit \(V_{out}\) is:
In the circuit shown in the figure, both OPAMPs are ideal. The output for the circuit \(V_{out}\) is:
Show Hint
- \(20V_1 + 10V_2\)
- \(-20V_1 + 10V_2\)
- \(10V_1 - 20V_2\)
- \(20V_1 - 10V_2\)
The Correct Option is D
Solution and Explanation
Step 1: Analyze the first OPAMP (inverting amplifier).
For the first OPAMP: - Input \(V_1\) passes through resistor \(R\). - Feedback resistor = \(10R\). Gain = \(-\dfrac{10R}{R} = -10\). Thus, the output of the first OPAMP is \[ V_{A} = -10V_1 \]
Step 2: Input to the second OPAMP.
The second OPAMP is also in inverting configuration: - Input \(V_A\) enters through \(5R\), and feedback resistor = \(10R\). Thus, gain = \(-\dfrac{10R}{5R} = -2\). This gives contribution from the first stage as \(+20V_1\).
Step 3: Add contribution from \(V_2\).
The second OPAMP also receives \(V_2\) through resistor \(R\). For this branch: gain = \(-\dfrac{10R}{R} = -10\). Therefore, the total output is \[ V_{out} = (20V_1) + (-10V_2) = 20V_1 - 10V_2 \]
Step 4: Conclusion.
Hence, the output voltage is \(20V_1 - 10V_2\).
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