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Four resistances are connected by an ideal battery of emf 15 V , the circuit is in steady-state then the current (in ampere) in wire AB is:

Correct answer

2

Step-by-step solution

R e q = 3 4 + 8 6 = 25 12 ⇒       i 0 = V R e q = 24   A i 1 = 3 4 × 24 = 18   A i 2 = 4 6 × 24 = 16   A Current in the branch AB ∆ i = 2   A

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