A new flashlight cell of emf 1.5 volts gives a current of 15 A when connected directly to an ammeter of resistance 0.040Ω. The internal resistance of the cell is:
Correct answer: C. 0.06 Ω
- A. 0.04 Ω
- B. 0.10 Ω
- C. 0.06 Ω
- D. 10 Ω
Explanation
The total resistance in the circuit is the sum of the ammeter's resistance (R) and the cell's internal resistance (r). Using Ohm's Law for the entire circuit, E = I(R+r). Solving for r gives r = (E/I) - R = (1.5V / 15A) - 0.04Ω = 0.06 Ω.
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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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