Two heaters A and B are connected in parallel across the supply voltage. Heater A produces 500 kJ in 20 minutes and B produces 1000 kJ in 10 minutes. The resistance of A is 100 Ω. If the same heaters are connected in series across the same voltage, the heat produced in 5 minutes will be:
Correct answer: B. 100 kJ
- A. 200 kJ
- B. 100 kJ
- C. 50 kJ
- D. 10 kJ
Explanation
First, calculate the power and resistance of each heater, then find the total power when they are in series across the same voltage. This series power is used to calculate the heat produced in 5 minutes (300 s), which correctly yields 100 kJ.
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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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