A heating element of resistance (at itsoperating temperature) 239 is connected t

Question

A heating element of resistance (at itsoperating temperature) 239 is connected to abattery of emf 646 V and unknown internal resistance r. It isfound that heat energy is being generated in the heater element ata rate of 74.0 W. What is the rate at which heat energy isbeing generated in the internal resistance of the battery?

Explanation / Answer

Total Resistance in the circuit = 239 + r Voltage = 646 V So current in the circuit = 646 / (239+r) We know , power =I2R So power in heater element = I2239 Putting the values 74.0=I2*239 6462/(239+r)2 = 74.0 / 239 =>646/(239+r) = 0.556 => 646 = 133 + 0.556r =>r = 922.7 Putting in equation of current , I = 646 / (239+r) =646 / 1161.7 = 0.556 We are required to find power in internal resistance Power = I2r Power = 0.5562*922.7 = 285.3 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.