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For the circuit shown below, calculate the value of $I_{z}$ .

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25 mA

0.15 A

0.1 A

0.05 A

answer is A.

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Detailed Solution

$I = \frac{50}{1000} = 50 mA$

$I_{L} = \frac{50}{2000} = 25 mA$

$I_{z} = I_{1000} - I_{2000}$

$= 50 - 25 = 25 mA$

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