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A circuit consists of a 15 V battery with internal resistance 2 Ω connected to two parallel branches: one with 4 Ω and 6 Ω in series, and the other with 8 Ω. The total current supplied by the battery is:

Correct answer: B. 2 A

  • A. 1 A
  • B. 2 A
  • C. 3 A
  • D. 4 A

Explanation

Series branch resistance = 4 + 6 = 10 Ω. Parallel with 8 Ω: 1/Rₑ𝑞 = 1/10 + 1/8 = 4/40 + 5/40 = 9/40, so Rₑ𝑞 = 40/9 Ω. Total resistance = 40/9 + 2 = 58/9 Ω. Current I = 15/(58/9) = 15 × 9/58 ≈ 2.33 A.

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About Current Electricity

Steady current relates charge flow to potential difference, resistance and resistivity, including how temperature changes a conductor's resistance. Sources with internal resistance have terminal voltage below their emf when delivering current, and maximum power output occurs under a specific load condition rather than at zero resistance.

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