If the op-amp in the circuit shown below is considerd as an ideal op-amp then the value of current supplied by the 15 V battery is

[Assuming $β$ of the transistor to be very large]

0.5 mA

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65.2 mA

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15 mA

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50.2 mA

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Solution

The correct option is D 50.2 mA
$V_{0} = 5 V$

$∴ I_{E} = I_{C} = \frac{5}{100} = 50 m A$

$I_{Z} = \frac{15 - 5}{50 k Ω} = 0.2 m A$

[Since, $β$ is very large ] $V_{z} = 5 V$

$∴ I_{n e t} = 50 + 0.2 = 50.2 m A$

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