Consider the voltage reference circuit shown in figure below. Assume the op-amp to be ideal, then the value of current $I_{z}$ is equal to

0.11 mA

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0.5 mA

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0.57 mA

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0.24 mA

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Solution

The correct option is B 0.5 mA

$I_{1} = \frac{10 - 6.8}{5.6 k Ω}$
$I_{1} = 0.571 m A$
Now, $I_{2} = \frac{6.8 - 0}{100 k Ω}$
$I_{2} = 0.068 m A$
$∴ I_{z} = I_{1} - I_{2}$
$= 0.571 - 0.068$
$I_{z} = 0.5 m A$

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