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Consider the cicutir shown in fig. the current `i_3` is equal to
image

1 Answer

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by (92.8k points)
 
Best answer
Suppose current through different paths of the circuit is as
follows.
After applying KVL for loop (1) and loop(2), we get
`28if_1 = -6 -8`
or `i_1 = -(1)/(2)`
And `54i_2 = -6 -12 or i_2 = - (1)/(3)A`
Hence `i_3 = i_1+i_2 = -(5)/(6)A`
image

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