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Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is 
A
0.5 A
B
0.75 A
C
zero
D
0.25 A
Detailed Solution
$D_1$ (in series with 10 $\Omega$) is forward biased and conducts; being ideal it has zero resistance.
$D_2$ (in series with 20 $\Omega$) is reverse biased and does not conduct.
So current flows only through the 10 $\Omega$ branch: $i = \frac{5}{10} = 0.5$ A
$D_2$ (in series with 20 $\Omega$) is reverse biased and does not conduct.
So current flows only through the 10 $\Omega$ branch: $i = \frac{5}{10} = 0.5$ A