Question: Q. 5. For the circuit shown here, find the current flowing through the $1 \Omega$ resistor. Assume that the two diodes $D_{1}$ and $D_{2}$ are ideal diodes.
A [SQP 2013]
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Solution:
Ans. Diode $D_{1}$ is forward biased, while diode $D_{2}$ is reverse biased. Hence the resistances of (ideal) diodes $D_{1}$ and $D_{2}$ can be taken as zero and infinity respectively. $\quad 1 / 2$ The given circuit can therefore be redrawn as shown in the figure.
$\therefore$ Using Ohm’s law,
$$ I=\frac{6}{(2+1)} \mathrm{A}=2 \mathrm{~A} $$
$\therefore$ Current flowing in the $1 \Omega$ resistor, is $2 \mathrm{~A}$.
