Answer / kishor mahale

input resistance of the opamp is very high, so current is
cannot be flow through opamp and it passes through feedback
path. it create ground at input but it is not a actual
ground so we assume that it is virtual ground.

Answer / sachin

In an opamp the input impedance is very high because of the
first stage which is dual input balanced output differential
amplifier. Due to this, negligible current flows into the
opamp. This suggests that the potential difference across
the input impedance is zero and if any one terminal is
grounded then the potential at the other end is zero.
However it is called virtual ground as ordinary ground has
zero potential but can sink infinite current whereas virtual
ground has zero potential and can sink zero current.

Answer / shobha

we know that open loop gain of an ideal op-amp is
infinity.This implies that the differential voltage is
zero.That is, voltage at the inverting terminal=voltage at
the non inverting terminal.If the non inverting terminal is
grounded ,it means that the inverting terminal is
grounded.In other words there is a virtual short cicuit
between the two terminals.,but no current flows through it
since there is no physical connection between the two.This
is virtual ground in op amp.

Answer / m.electronis

since in an op amp the input resistance is very very high
so at the input node no current flows in to he opamp
and ,it passes through feed back resistance, so at the
input we assume virtual ground

Answer / azhar

Since the resistance between inverting and non inverting
terminlas of op amplifier is very high. Therefore ideally
no current or very samll current can flow through these
termonals. if we model this we must have
I+ = 0 = I-
So this creats a ground which is not like the actual ground
but it is said to be the virtual ground.

Answer / moumita

input resistance of the opamp is very high, so current is
cannot be flow through opamp and it passes through feedback
path. it create ground at input but it is not a actual
ground so we assume that it is virtual ground.

Answer / usama

Since an ideal amplifier has infinite input impedance so practically a very negligible voltage exist there and no current flows inside but it is actually an approximation in order to make calculation easier.It is called as virtual ground as it close to zero value so as actual ground is zero.....hope u got it

Answer / kedas

as v know dat the input impedance of the opamp is ideally infinite.basically it means impedance between inverting and non-inverting terminals is very high.by this v can say that internally current passage between the terminals is ideally zero.basic concept here is current between two equal voltages is zero.using dis v can say dat voltages at two terminals are equalbcz current is zero.this is called virtual short.if any one terminal is grounded the other terminal is also grounded.simple very easy concept!!!!!!!!!!!!!!!!!!hope it helps:)

ur kedas!!!!!!!!!!!!!!!!!!!!!

Answer / p santosh kumar

Concept of virtual ground compraises of two things ,one is node voltage and current through op amp . input impedance of op amp is huge therfore no current can pass throughit ideally therefore current diverts through low resistance path and now as we know that for infinite impedance the difference of two inputs of op amp should be zero therefore one of the input should be assumed as grounded virtually but not physically thus comes the concept of virtual ground.
As I=0 v1=v2 , therefore if either of v1 or v2 is zero other input node to op amp is assumed to be zero

Answer / mohammad shariful islam moon

To achieve a reasonable voltage at the output (and thus
equilibrium in the system), the output supplies the
inverting input (via the feedback network) with enough
voltage to reduce the potential difference between the
inputs to micro volts. The non-inverting (+) input of the
operational amplifier is grounded; therefore, its inverting
(-) input, although not connected to ground, will assume a
similar potential, becoming a virtual ground.

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