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(B) When a positive voltage of $8 \ V$ is applied,the green $LED$ is forward-biased and conducts,while the red $LED$ is reverse-biased.The current $I$

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$300$

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(B) When a positive voltage of $8 \ V$ is applied,the green $LED$ is forward-biased and conducts,while the red $LED$ is reverse-biased.The current $I$ flowing through the circuit is given by Ohm's law:$I = \frac{V_{source} - V_{LED}}{R}$Given $I = 20 \ mA = 20 	imes 10^{-3} \ A$,$V_{source} = 8 \ V$,and $V_{LED} = 2 \ V$:$20 	imes 10^{-3} = \frac{8 - 2}{R}$$R = \frac{6 \ V}{20 	imes 10^{-3} \ A} = \frac{6000}{20} \ \Omega = 300 \ \Omega$The reverse voltage across the red $LED$ is $2 \ V$,which is less than its reverse breakdown voltage of $3 \ V$,so it is safe.

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