question_answer

Consider the circuit shown in the figure. The current \[{{I}_{3}}\] is equal to                                        [AMU 1995]

A)                    5 amp

B)                    3 amp

C)                    \[-3\,amp\]

D)                    \[-5/6\,amp\]

Correct Answer: D

Solution :

           Suppose current through different paths of the circuit is as follows. After applying KVL for loop (1) and loop (2) We get \[28{{i}_{1}}=-6-8\]   Þ \[{{i}_{1}}=-\frac{1}{2}A\] and      \[54{{i}_{2}}=-6-12\] Þ \[{{i}_{2}}=-\frac{1}{3}A\] Hence \[{{i}_{3}}={{i}_{1}}+{{i}_{2}}=-\frac{5}{6}A\]