For the circuit shown below.

the output voltage $V_{0}$ is ___V.
6

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Solution

The correct option is A 6

Due to virtual ground,
$V_{A} = 1 V$
$V_{C} = 2 V$

KCL at node-A,

$\frac{0 - 1}{2 k Ω} = \frac{1 - V_{B}}{2 k Ω} + \frac{1 - 2}{1 k Ω}$

$- 1 = 1 - V_{B} - 2$
$V_{B} = 0 V$

KCL at node-C,

$\frac{V_{B} - V_{C}}{2 k Ω} + \frac{V_{A} - V_{C}}{1 k Ω} = \frac{V_{C} - V_{0}}{2 k Ω}$

$\frac{0 - 2}{2 k Ω} + \frac{1 - 2}{1 k Ω} = \frac{2 - V_{0}}{2 k Ω}$

$\frac{- 1}{1 k Ω} - \frac{1}{1 k Ω} = \frac{1}{1 k Ω} - \frac{V_{0}}{2 k Ω}$

$\frac{V_{0}}{2 k Ω} = \frac{3}{1 k Ω}$

$V_{0} = 6 V$

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