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Network Theory Circuit

What will be v1(0) and vo(0)? As the voltage across the capacitor cannot change abruptly im sure that v(0) will be 0 as v(t) for t<0 is 0. But im confused about v1(0) and vo(0). Will they be 0 as well?

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  • \$\begingroup\$ parasitic capacitors exist at every node (shown and un-shown) hence V1 will be zero at the instant the switch closes. \$\endgroup\$ – Andy aka Jun 19 '17 at 9:22
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Assumption : the opamp is ideal

As soon as the switch is closed, the voltage V1 will have a value = 2V. As per the law of virtual ground, the voltage at -ve end of opamp will be equal to 2 volts as well. All this is happening at t=0 and at t=0, the capacitor behaves as a short circuit. So, Vo will be equal to the voltage at -ve end of the opamp (as the capacitor is short now) which will make Vo = 2V as well.

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  • \$\begingroup\$ Great ! Welcome \$\endgroup\$ – MoHit Jun 19 '17 at 9:12
  • \$\begingroup\$ As a side note, what will actually happen is that V1 will rise very fast to 2 V, Vo will follow, then the opamp will answer with its own frequency response changing Vo. \$\endgroup\$ – Vladimir Cravero Jun 19 '17 at 9:41
  • \$\begingroup\$ @VladimirCravero Rightly said. \$\endgroup\$ – MoHit Jun 19 '17 at 9:46
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Depending on the programm and Settings you are using.

Since the resistors may be ideal, so have no time derivative aspect its depending on the voltage source.

If your set up the voltage source as 3V initial, you will have V1(0) = 2V

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  • \$\begingroup\$ Yes you are correct, taking v1(0) = 2v matches the answer. Thank You \$\endgroup\$ – Chirag Gupta Jun 19 '17 at 9:01

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