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In the network shown in the figure, the diodes $D_{1}$ has forward resistance of $10 Ω$ and reverse resistance of $200 Ω$ . whereas diode $D_{2}$ is ideal .Then current flowing through the cell of emf 10 volt and internal resistance 2 ohm is

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1 A

1/4 A

1/3 A

1/2 A

answer is C.

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Detailed Solution

$D_{1}$ is forward biased and $D_{2}$ does not conduct.

$∴ i = \frac{10}{(8 + 2 + 10 + 10)} A = \frac{1}{3} A$

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