Q9 / 237 · NTA Abhyas (NEET) 2020 · Physics · Semiconductors

Current in the given circuit will be

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

Solution

The diode in the lower branch is forward and diode in the upper branch is reversed biased, so

i = \frac{5}{20 + 30} = \frac{5}{50} A

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

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