Consider the ideal op-amp circuit shown in the figure below:

$The value of output voltage V_{0} is equal to$

0 V for V_{1} = 3 V and V_{2} = - 3 V

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- 1.5 V for V_{1} = 1 V and V_{2} = - 3 V

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0.75 V for V_{1} = 2 V and V_{2} = - 3 V

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- 4.5 V for V_{1} = - 5 V and V_{2} = - 1 V

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Solution

The correct option is D $- 4.5 V for V_{1} = - 5 V and V_{2} = - 1 V$

$Applying KCL at node V_{+} we get,$

$\frac{V^{+} - V_{1}}{R} + \frac{V^{+} - V_{2}}{R} + \frac{V^{+}}{\frac{R}{2}} = 0$

$4 V^{+} = V_{1} + V_{2}$

$V^{+} = \frac{V_{1} + V_{2}}{4}$

$Now, V_{0} = (1 + \frac{2 R}{R}) V^{+}$

$Therefore, V_{0} = (1 + \frac{2 R}{R}) . (\frac{V_{1} + V_{2}}{4}) = \frac{3}{4} (V_{1} + V_{2})$

$For V_{1} = 3 V and V_{2} = - 3 V, V_{0} = 0 V$

$For V_{1} = 1 V and V_{2} = - 3 V, V_{0} = - 1.5 V$

$For V_{1} = 2 V and V_{2} = - 3 V, V_{0} = - 0.75 V$

$For V_{1} = - 5 V and V_{2} = - 1 V, V_{0} = - 4.5 V$

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