The given circuit has two ideal diodes $D_1$ and $D_2$ connected as shown in the figure. The current flowing…
The given circuit has two ideal diodes $D_1$ and $D_2$ connected as shown in the figure. The current flowing through the resistance $R_1$ will be
$2 \mathrm{~A}$
$3.3 \mathrm{~A}$
$2.5 \mathrm{~A}$
$7 \mathrm{~A}$
Solution
The ideal diode $D_1$ is reversed biased whereas ideal diode $D_2$ is forward biased. Thus, $D_1$ acts as an open switch while $D_2$ as a closed switch as shown in the circuit.
Thus, current flowing through $R_1, I=\frac{V}{R_1+R_3}=\frac{10}{2+2}=2.5 \mathrm{~A}$
