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Question

Assuming that the op-amp in the circuit shown is ideal. The output voltage V0 is

A
3V1+112V2
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B
32V1+72V2
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C
52V13V2
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D
2V152V2
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Solution

The correct option is A 3V1+112V2



By applying KCL at node 1 we get ,

V2V1R+V22R+V2V03R=0

6V26V1+3V2+2V22V06R=0

11V26V12V0=0

2V0=6V1+11V2

V0=3V1+112V2

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