For the given circuit shown below, the diode has $V_{o n} = 0.7$ V, but otherwise acts as ideal diode. The magnitude of current $i_{2}$ (in mA) is equal to ______.
(Answer upto 2 decimal places)

0.25

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Solution

The correct option is A 0.25
For given circuit,

Let the diode be ON,

$V_{A} = 2 - 0.7 = 1.3 V$

So, $i_{2} = \frac{1.3}{6 k Ω} = 0.2166 m A$

$i_{1} = \frac{0.7}{2 k Ω} = 0.35 m A$

$i_{d} = i_{2} - i_{1} = - v e$ (not possible case)

If the diode is OFF,

$i_{2} = \frac{2 V}{2 k Ω + 6 k Ω} = 0.25 m A$

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