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The circuit shown below contains two ideal diodes, each with a forward resistance of $50\ \Omega$. If the battery voltage is 6 V, the current through the $100\ \Omega$ resistance (in Amperes) is:
(1) 0.036
(2) 0.020
(3) 0.027
(4) 0.030
$D_2$ is reverse biased. $R_{eq} = 150 + 50 + 100 = 300\ \Omega$. $I = 6/300 = 0.02$ A.
BETA
