In the circuit shown, three identical ammeters A1, A2, and A3 are connected. If A1 reads 5 A and A3 reads 13 A, what will be the reading of A2 according to Kirchhoff's Current Law (KCL)?

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:This problem checks understanding of Kirchhoff's Current Law (KCL) for parallel branches measured with ideal ammeters. In such networks, the current shown by an upstream meter equals the algebraic sum of branch currents measured by downstream meters. Given Data / Assumptions: A1 indicates 5 A. A3 indicates 13 A. All ammeters are identical and ideal (negligible internal resistance). A2 is placed so that A3 measures the total of A1 and A2. Concept / Approach:Kirchhoff's Current Law states that the sum of currents entering any node equals the sum leaving that node. If A3 measures the total current after the junction where branches with A1 and A2 recombine, then I3 = I1 + I2. Rearranging gives I2 = I3 − I1. Step-by-Step Solution: Apply KCL at the node: I3 = I1 + I2.Substitute values: 13 = 5 + I2.Solve: I2 = 13 − 5 = 8 A. Verification / Alternative check: Check that 5 A + 8 A = 13 A, satisfying KCL. Why Other Options Are Wrong: 13 A, 18 A, or 12 A do not satisfy I3 = I1 + I2 with the given numbers.10 A is a distractor that also violates KCL for the stated readings. Common Pitfalls: Mixing up series and parallel current relationships or assuming ammeters alter the current. Final Answer: 8 A

In the circuit shown, three identical ammeters A1, A2, and A3 are connected. If A1 reads 5 A and A3 reads 13 A, what will be the reading of A2 according to Kirchhoff's Current Law (KCL)?

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