Q83 / 120 · VITEEE 2014 · Physics

Current in the circuit will be Figure

  • A
    $\frac{5}{40} \mathrm{~A}$
  • Correct answer
    B
    $\frac{5}{50} \mathrm{~A}$
  • C
    $\frac{5}{10} \mathrm{~A}$
  • D
    $\frac{5}{20} \mathrm{~A}$

Solution

Here, diode in lower branch is forward and in upper branch is reversed biased $\therefore \quad \mathrm{i}=\frac{5}{20+30}=\frac{5}{50} \mathrm{~A}$

Hence, the correct answer is option: 'B'.

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