If an ideal diode is used in the given circuit, find the current through each resistance

Options

A. `I_(1)=(9)/(6)A, I_(2)=3A`
B. `I_(1)=3A, I_(2)=3/2 A`
C. `I_(1)=2/3 A, I_(2)=1/3 A`
D. `I_(1)=1/3 A, I_(2)=2/3 A`

Answer

A

Explanation

`I=(E)/(R+r)=(9)/(2+r)=9/2 A`
Now `I_(1)=((R_(2))/(R_(1)+R_(2)))I=((3)/(6+3))(9)/(2)=3/2=9/4A`
and `I_(2)=((R_(1))/(R_(1)+R_(2)))I=((6)/(6+3))9/2 =3A`