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Q.
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit ?
KCETKCET 2015Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
In the given circuit, $D_{1}$ is forward biased and $D_{2}$ is reverse biased. So, $D_{1}$ works like a closed switch and $D_{2}$ works like an open circuit. So, circuit becomes
Now, current in the circuit, $I=\frac{E}{R_{s}}=\frac{12}{4+3}$
$I=\frac{12}{7}$
$I=1.71\, A$