The correct option is C 2 A
Let current I3 is drawn from the battery of 12 V.
This is divided into 2 streams of current I1 and I2.
With help of kirchhoff's law on junction E
I3 = I1 + I2
Using the loop law,
For the loop EADFE, starting from a 12 V battery and going anti-clockwise.
Loop EADFA
Applying KVL , from 12V battery
12−5−2I1−3(I1+I2)=0
7−2I1−3I1−3I2=0
7−5I1−3I2=0
5I1+3I2=7 ……..(1)
Using the loop law,
For the loop EBCFE, starting from a 12 V battery and going clockwise.
Loop EBCFE
Applying KVL , from 12V battery
12−4I2−3(I1+I2)=0
12−4I2−3I1−3I2=0
12−3I1−7I2=0
3I1+7I2=12 …….(2)
Solving eq (1) and eq(2)
3(5I1+3I2=7)=>15I1+9I2=21
5(3I1+7I2=12)=>15I1+35I2=60
−26×I2=−39
I2=1.5 A putting this in eq (1)
5I1+3×1.5=7
5I1=2.5
I1=0.5 A
I3=I1+I2=0.5 A+1.5 A=2 A