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

then the value of current gain i.e., $\frac{I_{o}}{I_{i}}$ for the circuit is equal to

(1 - \frac{R_{1}}{R_{2}})

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(\frac{R_{1}}{R_{2}})

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(1 + \frac{R_{1}}{R_{2}})

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(1 + \frac{R_{2}}{R_{1}})

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Solution

The correct option is C $(1 + \frac{R_{1}}{R_{2}})$

$I_{0} = I_{1} + I_{2}$
Now, $V_{x} = - I_{1} R_{1}$
$∴ I_{2} = \frac{0 - V_{x}}{R_{2}} = \frac{I_{1} R_{1}}{R_{2}}$
$∴ I_{0} = I_{1} + \frac{R_{1}}{R_{2}} I_{1}$
$= (1 + \frac{R_{1}}{R_{2}}) I_{1} = (1 + \frac{R_{1}}{R_{2}}) I_{i}$

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