Question
Download Solution PDFFor the ideal Op-Amp circuit shown in the figure, V0 is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFAfter re-drawing the circuit, we get
By applying KCL at node ‘A’, we get
\(\frac{{1 - {V_A}}}{{1k}} = \frac{{{V_A}}}{{1k}} \Rightarrow {V_A} = \frac{1}{2}\;Volt\)
Now applying KCL at node ‘B’, we get
\(\frac{{1 - V - B\;}}{{1k}} = \frac{{{V_B} - {V_0}}}{{2k}}\)
2 – 2 VB = VB = V0
Where, VB = VA = 0.5 volt (∵ virtual ground)
∴ 2 – 1 = 0.5 – V0
⇒ V0 = -0.5 Volt
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