# The op-amp in the circuit shown in the figure works in linear mode. The output voltage Vo is

The op-amp in the circuit shown in the figure works in linear mode. The output voltage Vo is

#### 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+ − Vo = 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 − Vo = 0

Vo = 6V

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