SemiconductorMedium
Question
The current / in the circuit shown below is:
(All diodes are ideal and identical)
Options
A.$\frac{5}{3}\text{ }A$
B.$\frac{5}{9}\text{ }A$
C.$\frac{1\ \ }{3}\text{ }A$
D.$\frac{15}{2}\text{ }A$
Solution
Sol. [Key Concept: Ideal-diode circuit — forward R=0, reverse R=∞] For ideal diode, forward resistance $= 0$ and reverse biased resistance $= \infty$
∴ Circuit can be redrawn as,
$$\begin{matrix} I & \ = \frac{10}{2} + \frac{10}{4} \\ & \ = \frac{30}{4} = \frac{15}{2}\text{ }A \end{matrix}$$
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