The current (I) in the circuit will be :

\frac{5}{40} A

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\frac{5}{50} A

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\frac{5}{10} A

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\frac{5}{20} A

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Solution

The correct option is C $\frac{5}{50} A$
No current will pass through the upper circuit because diode is reversed biased.
Hence current will only flow through the lower circuit
$R_{e f f} = 20 + 30 = 50 Ω$
$I = \frac{V}{R_{e f f}} = \frac{5}{50} Ω$

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