Question

An op-amp with an open loop gain of $1000V/V$ is used in the inverting configuration. If in this application, the output voltage ranges from $+10Vto-10V$, then the maximum voltage by which the virtual ground node departs from its ideal value is equal to

• A. $+10V$ from ideal voltage of zero
  
• B. $+10V$ from ideal voltage of $V_0$
• C. $+10mV$ from ideal voltage of zero
  
• D. $+10mV$ from ideal voltage of $V_0$

Answer 1

The correct option is C $+10mV$ from ideal voltage of zero


For amplifier to have a valid

$V_{out}=A(V_1-V_2)$

$(V_1-V_2)=\frac{V_{out}}{A}$

now, for vitural ground $A\to \infty$
But if $A\ne \infty$, then

$V_t=\frac{\pm V_{out}}{A}=\frac{\pm 10}{1000}=\pm 10mV$