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Tardigrade
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Physics
Current in the circuit will be
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Q. Current in the circuit will be
AIPMT
AIPMT 2001
Semiconductor Electronics: Materials Devices and Simple Circuits
A
$\frac{5}{40} A$
12%
B
$\frac{5}{50} A$
62%
C
$\frac{5}{10} A$
16%
D
$\frac{5}{20} A$
10%
Solution:
The diode in lower branch is forward and diode in upper branch is reversed biased,
so $i=\frac{5}{20+30}=\frac{5}{50} A$
