In the following circuit,the currents i_1 and | vedclass

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(D) In the given circuit,the first diode is reverse-biased because the potential at its anode is lower than the potential at its cathode (due to the $

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$0, 5\, mA$

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(D) In the given circuit,the first diode is reverse-biased because the potential at its anode is lower than the potential at its cathode (due to the $14\, k\Omega$ resistor path to ground). Therefore,no current flows through the first diode branch.Consequently,the circuit simplifies to a series combination of the $10\, V$ source and the $2\, k\Omega$ resistor. The current $i$ flowing through this branch is:$i = \frac{10\, V}{2\, k\Omega} = 5\, mA$.Since the first diode is open-circuited,$i_1 = 0$.The current $i_2$ flows through the second diode,which is forward-biased. Thus,$i_2 = i = 5\, mA$.Therefore,$i_1 = 0$ and $i_2 = 5\, mA$.

In the following circuit,the currents $i_1$ and $i_2$ are respectively:

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