In the given figure, each diode has a forward bias resistance of 30 Ω and infinite resistance in reverse bias. The current I 1 will be :

Step-by-step solution

We know that in reverse bias ideal diode does not pass current it only works in forward bias so, the current in the third branch is zero, it means the net resistance of the upper two branch is, 1 R e q = 1 160 + 1 160 ⇒ R e q = 80   Ω ⇒ R ' e q = 80 + 20 = 100   Ω ⇒ i = 200 100 = 2   A