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(A) In the given circuit,the diode is connected in reverse bias because the positive terminal of the battery is connected to the $n$-side (cathode) an

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$0.0 \ A, 0.2 \ A$

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(A) In the given circuit,the diode is connected in reverse bias because the positive terminal of the battery is connected to the $n$-side (cathode) and the negative terminal is connected to the $p$-side (anode) of the diode.Since the diode has infinite reverse bias resistance,the current $I_1$ flowing through the diode branch is $I_1 = 0.0 \ A$.The resistor of $50 \ \Omega$ is connected in parallel to the battery of $10 \ V$. Therefore,the current $I_2$ flowing through the resistor is given by Ohm's law:$I_2 = \frac{V}{R} = \frac{10 \ V}{50 \ \Omega} = 0.2 \ A$.Thus,$I_1 = 0.0 \ A$ and $I_2 = 0.2 \ A$.

If a diode having infinite reverse bias resistance is connected in a circuit as shown in the figure,then $I_1$ and $I_2$ are respectively:

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