In the figure shown, the current in the 10 V battery is close to :

KJR, i=i1+i2​KVL in left loop, 20−5i−10i1−2i=0​⇒i1=20−7i10KVL in outermost loop, 20−5i−10−4i2−2i=0​⇒i2=10−7i4using first equation , ​i=20−7i10+10−7i4⇒40i=80−28i+100−70i​⇒i=180138A∴i2=10-71801384=0.21A