SOLUTION

According to the virtual ground concept for inverting amplifier

• The non-inverting input (+) is connected to the 0 V line.
• Part of the output voltage (or signal) is connected to the inverting input (—)
• The input voltage (or signal) is connected to the inverting input.
• A positive voltage is equal to the negative voltage

V⁺ = V⁻

Applying Nodal analysis on the Positive terminal i.e V⁺

$\begin{array}{l}\dfrac{{{V^ + } – 0}}{{20K\Omega }} + \dfrac{{{V^ + } – {V_{out}}}}{{100K\Omega }} = 0\\\\= \dfrac{{5{V^ + } + {V^ + } – {V_{out}}}}{{100K\Omega }}\end{array}$

6V⁺ − V_o = 0 —–1

Applying Nodal analysis on the Negative terminal i.e V⁻

$\begin{array}{l}\dfrac{{{V^ – } – 1.2}}{{20K\Omega }} + \dfrac{{{V^ – } – 0}}{{100K\Omega }} = 0\\\\= \dfrac{{5{V^ – } + {V^ – } – 6}}{{100K\Omega }}\end{array}$

6V⁻ = 6

V⁻ = 1 V

Since V⁺ = V⁻

V⁺ = V⁻ = 1 V

Putting the value of V⁺ = 1 V in equation 1

6 × 1 − V_o = 0

V_o = 6V