Chapter 26, Problem 053 A 120 V potential difference is applied to a space heater that...

Question

Question

Chapter 26, Problem 053 A 120 V potential difference is applied to a space heater that dissipates 708 W during operation. (a)

Answer 1

Answer #1

a) P = V²/R

Therefore , R = V²/P = 120²/708 = 20.34 ohm

b) i = V/R = 120/20.34 = 5.9 A = 5.9 C/s

Now,

1.6 × 10^-19 C = 1 electron

1 C = 1/(1.6×10^-19) = 6.25 × 10¹⁸ electron

5.9 C = 5.9 × 6.25 × 10¹⁸ = 3.69 × 10¹⁹ electron

Electron flow rate = 3.69 × 10¹⁹ electrons/sec

Answer 2

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