Analog electronics circuits miscellaneous
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Find Av = v0 vi
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VA = Vi (due to virtual ground) KCL at node A.
VA – VB + VA – 0 = 0 R R
or 2 VA = VB
KCL at node B.VB – VA + VB + VB – V0 = 0 R R R or 2VA – VA + 2VA + 2VA – V0 = 0 (VB = 2 VA) R R R or Vi + 2Vi + 2Vi = V0 (Vi = VA) R R R R
or 5 Vi = V0or V0 = 5 Vi
Correct Option: A
VA = Vi (due to virtual ground) KCL at node A.
VA – VB + VA – 0 = 0 R R
or 2 VA = VB
KCL at node B.VB – VA + VB + VB – V0 = 0 R R R or 2VA – VA + 2VA + 2VA – V0 = 0 (VB = 2 VA) R R R or Vi + 2Vi + 2Vi = V0 (Vi = VA) R R R R
or 5 Vi = V0or V0 = 5 Vi
- In a time base generator a constant current source (I) is used to drive a capacitance (C). The output voltage across the capacitance as a function of time (t) is given by:
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q = C V
It = C Vor V = It C Correct Option: A
q = C V
It = C Vor V = It C
- A differentiator is rarely used in analog computers as it:
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NA
Correct Option: C
NA
- The circuit shown below is best described as:
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NA
Correct Option: C
NA
- In the circuit, shown below I0 is given by:
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Due to virtual ground voltage at terminal is Vs. Now, KCL at A
Io – Vs = 0 Rs or Io = Vs Rs
Correct Option: B
Due to virtual ground voltage at terminal is Vs. Now, KCL at A
Io – Vs = 0 Rs or Io = Vs Rs