Assume ideal OP-Amp. The input impedance (Ri) of an OP-AMP is ideally Infinite. Hence current “I” flowing from one input terminal to other will be zero as shown .Thus Voltage drop across $R_i=0$ Both the i/p terminals will be at the same potential. → i.e. virtually shorted

i.e. $V_0= Av.Vd$ → differential i/p vtg.

$V_0$ = o/p Vtg

Vd=Differential Input vtg

Av = open loop gain

$∴Vd=\frac{V_0}{Av}$

But Av=∞ for an ideal op-amp and $Av = 2 \times 10^5$ for IC 741

For ideal, Vd = 0, Thus potential difference between input terminals = 0

When we short cicuit two pts, they will have same potential. Due to this reason , the two OP-AMP terminals which are almost equipotential are to be virtually(not actually) short circuited.

Virtual Ground :

If (+) terminal is connected to grid, they due to “virtual short”, (-) terminal will also be grid potential. Hence it is “virtual ground”

Similarly if (-) terminal is connected to grid then (+) terminal will be at “virtual grid” potential.

The concept of virtual ground has been used extensively in analyzing various closed loop configuration, especially we use this concept in the inverting amplifier analysis.