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Question

For the given circuit below it is given that
Vin(t)=102cos(2π60t) V and V0(t)=102cos(2π60t120) V.
(Assuming the OPAMP to be ideal)
The possible value of R is

A
35.94 kΩ
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B
40.00 kΩ
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C
44.94 kΩ
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D
45.94 kΩ
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Solution

The correct option is D 45.94 kΩ
VP=1sC1sC+R.Vin=11+sRC.VinDue to virtual ground,VN=VP=11+sRC.Vin

Apply KCL,

VinbVN100 k=VNV0100k

V0=2VNVin

V0=21+sRC.VinVin

V0Vin=1sRC1+sRC

Put (s=jω)

V0(jω)Vin(jω)=1jωRC1+jωRC

V0(jω)Vin(jω)=tan1ωRCtan1ωRC=2tan1ωRC

V0(jω)Vin(jω)=120=2tan1ωRC

ωRC=tan 60

R=tan602π×60×0.1×106=45.94 kΩ

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