Consider the circuit shown in the figure. The current i3 is equal to
A
5 amp
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B
3 amp
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C
– 3 amp
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D
−56 amp
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Solution
The correct option is D−56 amp Suppose current through different paths of the circuit is as follows. After applying KVL for loop (1) and loop (2) We get 28i1=−6−8⇒i1=−12A and 54i2=−6−12⇒i2=−13A Hence i3=i1+i2=−56A