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Question

Consider the circuit shown in the figure below:

Then,

A
the output istR2C2u(t) for Vin(t)=δ(t)
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B
the output ist2R2C2u(t) for Vin(t)=u(t)
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C
the output is1R2Cu(t) for Vin(t)=tu(t)
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D
the output ist2R2C2u(t) for Vin(t)=u(t)
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Solution

The correct option is B the output ist2R2C2u(t) for Vin(t)=u(t)
Taking Laplace transform, we have,


I1=Vin2R1+sRC and If(s)=Vout(s)×sRC22(1+sRC)


Now, I1(s)=If(s)

Vin(s)2R(sRC+1)=RC2s22(sRC+1)VVout(s)

Vout(s)=1s2R2C2Vin(s)=1R21C2s3=1R21C2s3

Vout(s)=t22R2C2u(t)

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