Consider the circuit shown in the figure. The current `I_(3)` is equal to

Options

A. `5 amp`
B. `3 amp`
C. `- 3 amp`
D. `- 5//6 amp`

Answer

D

Explanation

(d) Suppose current through paths of the circuit is as follows.
After applying `KVL` for loop (1) and loop (2)
we get `28i_(1) = - 6 - 8 implies i_(1) = - (1)/(2) A`
and `54i_(2) = - 6 - 12 implies i_(2) = - (1)/(3) A`
Hence `i_(3) = i_(1) + i_(2) = - (5)/(6) A`