Assuming that the Op-amp in the circuit shown is ideal, V o is gi
Assuming that the Op-amp in the circuit shown is ideal, Vo is given by
A. <span class="math-tex">\(\frac{5}{2}{V_1} - 3{V_2}\)</span>
B. <span class="math-tex">\(2{V_1} - \frac{5}{2}{V_2}\)</span>
C. <span class="math-tex">\(- \frac{3}{2}{V_1} + \frac{7}{2}{V_2}\)</span>
D. <span class="math-tex">\(- 3{V_1} + \frac{{11}}{2}{V_2}\)</span>
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
An ideal op-amp follows the virtual ground concept, i.e. the positive and negative terminals of the Op-amp remains at the same potential:
V+ = V-
Also, the Inputs terminals current of an ideal opamp is zero.
Calculation:
Using KCL at the negative terminal of the Opamp, we can write:
\(\frac{{{V_1} - {V_2}}}{R} = \frac{{{V_2}}}{{2R}} + \frac{{{V_2} - {V_0}}}{{3R}}\)
On solving it, we get:
\({V_{out}} = \frac{{11}}{2}{V_2} - 3{V_1}\)