# NEET Questions Solved

Consider the circuit shown in the figure. The current I3 is equal to

(1) 5 amp

(2) 3 amp

(3) –3 amp

(4) –5/6 amp

(4) Suppose current through different paths of the circuit is as follows.

After applying KVL for loop (1) and loop (2)

We get $28{i}_{1}=-6-8$${i}_{1}=-\frac{1}{2}A$

and $54{i}_{2}=-6-12$${i}_{2}=-\frac{1}{3}A$

Hence, ${i}_{3}={i}_{1}+{i}_{2}=-\frac{5}{6}A$

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