The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be

3.13 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.5 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10.0 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.43 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2.5 A$
From figure, 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 equivalent circuit.
Thus current flowing through $R_{1}$ , $I = \frac{V}{R_{1} + R_{3}} = \frac{10}{2 + 2} = 2.5 A$

Suggest Corrections

Similar questions

R_{1}

R_{1}

50 Ω

Join BYJU'S Learning Program