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

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0.42 A from positive to negative terminal

0.21 A from positive to negative terminal

0.36 A from negative positive terminal

0.71 A from positive to negative terminal

answer is B.

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Detailed Solution

KJR, i=i1+i2KVL 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

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