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A circuit with a 10 V battery and three resistors (2 Ω, 3 Ω, 5 Ω) in series is shunted by a 4 Ω resistor across the 3 Ω resistor. The current through the 4 Ω resistor is:

Correct answer: A. 0.5 A

  • A. 0.5 A
  • B. 1 A
  • C. 1.5 A
  • D. 2 A

Explanation

Series resistance without shunt = 2 + 3 + 5 = 10 Ω. With 4 Ω shunting 3 Ω, the equivalent resistance of parallel = (3 × 4)/(3 + 4) = 12/7 Ω. Total resistance = 2 + 12/7 + 5 = 61/7 Ω. Total current I = 10/(61/7) = 70/61 A. Voltage across parallel = (70/61) × (12/7) ≈ 2 V. Current through 4 Ω = 2/4 = 0.5 A.

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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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