The magnitude and direction of the current in the following circuit is [Circuit has two batteries: 1

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Question

The magnitude and direction of the current in the following circuit is [Circuit has two batteries: 10V and 5V with 10Ω resistance]

Accepted Answer

Correct Answer: B — 0.5 A from A to B through EStep-by-step Solution:Applying Kirchhoff's Voltage Law (KVL) around the loop:Net EMF = 10 V − 5 V = 5 V (assuming batteries are in opposition)Total resistance = 10 ΩCurrent I = Net EMF / Total resistance = 5/10 = 0.5 ADirection: from higher to lower potential, flowing from A to B through the external circuit (through E).Why other options are wrong:A (0.2 A): Would require net voltage = 2 V.C (5/9 A ≈ 0.56 A): Not obtained from these values.D (1.5 A from B to A): Both magnitude and direction are incorrect.Key Concepts:Kirchhoff's Voltage Law (KVL): ΣV = 0 around any closed loop. Sum of EMFs = Sum of IR drops.Kirchhoff's Current Law (KCL): Sum of currents entering a node = Sum of currents leaving.For opposing EMFs: Net EMF = E₁ − E₂; current I = (E₁ − E₂)/Rtotal

The magnitude and direction of the current in the following circuit is [Circuit has two batteries: 10V and 5V with 10Ω resistance]

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The magnitude and direction of the current in the following circuit is [Circuit has two batteries: 10V and 5V with 10Ω resistance]

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