Consider the circuit shown in the figure. The current ${I_3}$ is equal to
Diffcult
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(d)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 ${i_3} = {i_1} + {i_2} = - \frac{5}{6}\,A$
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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