The currents in the three branches of a parallel circuit are 3A, 4A and 5A. What is the current leav

Question

The currents in the three branches of a parallel circuit are 3A, 4A and 5A. What is the current leaving it?

Examoogle · Accepted Answer

12 A

Explanation: Given: Branch currents in a parallel circuit are 3A, 4A, and 5A.

Formula Used: $$I_{total} = I_1 + I_2 + I_3$$

Steps: Identify the Principle: According to Kirchhoff's Current Law (KCL), the sum of currents entering a junction must equal the sum of currents leaving the junction.. Apply KCL: Since all three branches are in parallel, the current leaving the circuit is the algebraic sum of the currents flowing through the branches.. Calculate the Result: Substituting the given values $I_1 = 3A$, $I_2 = 4A$, and $I_3 = 5A$ into the equation gives $3 + 4 + 5 = 12A$.

Answer: D is correct because by Kirchhoff's Current Law, the total current leaving a parallel combination of branches is the sum of the individual branch currents, which equals 12A.

Exam Tip: KCL is the fundamental basis for nodal analysis in circuit theory; remember that it holds true for any junction regardless of the complexity of the components.

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