Question
The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R _1$ will be
✓
Answer
$(a) 2.5 A$
Explanation: $D_1$ is reverse biased and $D_2$ is forward biased. $D_1$ blocks current.
Hence, Current will flow through $10 V$ cell, $R _1, D _2$ and $R _3$.
$\therefore I =\frac{\varepsilon}{R_1+R_3}$
$=\frac{10 V}{(2+2) \Omega}$
$=2.5 A$
Explanation: $D_1$ is reverse biased and $D_2$ is forward biased. $D_1$ blocks current.
Hence, Current will flow through $10 V$ cell, $R _1, D _2$ and $R _3$.
$\therefore I =\frac{\varepsilon}{R_1+R_3}$
$=\frac{10 V}{(2+2) \Omega}$
$=2.5 A$
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