A heater consists of a variable resistor (a resistor whose resistance can be var

Question

A heater consists of a variable resistor (a resistor whose resistance can be varied) connected across an ideal voltage supply. (An ideal voltage supply is one that has a constant emf and a negligible internal resistance.) To increase the heat output, should you decrease the resistance or increase the resistance? increase the resistance decrease the resistance Explain your answer. This answer has not been graded yet. A 10 gauge copper wire carries a current of 17 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm^2.) 0.23 mm/s

Explanation / Answer

1. as the voltage source is ideal,
   let the resistance of the resistor be R
   then from joules heating we know
   Heat, H = V^2t/R
   so, reducing the Resistance increases the heat dissipated by the heeater for a constant voltage
2. given, 10 gauge copper wire
   current, i = 17 A
   cross section of wire, A = 5.261*10^-6 m^2
   the formula for drift velocity is
   u = i/nqA
   now there is 1 free electron per atom of copper
   density of copper , rho = 8960 kg/m^3
   mass number of copper = 63.546 u
   so, 1 mol Cu has 63.546 g = 0.063546 kg
   1 kg cu = 1/8960 m^3 copper
   so number density n = 6.62*10^23*8960/0.063546 = 9.334*10^28 electrons/m^3
   q = charge on electron = 1.6*10^-19 C
   so, u = 17/nqA = 2.1636*10^-4 m/s = 0.21636 mm/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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