EET 102
Fall 1997
Prof. Ken Reid
9/17/97

HW 3 - Answers

ch.4 # 16) Graph of current-voltage characteristics of a 2k ohm and 50 k ohm resistor:
(Note - you could graph V vs. I or I vs. V)

R = 2k ohm

R = 50k ohm
V (V)
I (mA)
V (V)
I (mA)
0 V 0 mA 0 V 0 mA
10 V 5 mA 250 V 5 mA
20 V 10 mA 500 V 10 mA
40 V 20 mA 1 kV 20 mA

ch 4 # 21a) Find the energy (in Joules):

W = P * t ... (where t is time in seconds)
W = 2W * 8 hours * (60 min/hr) * (60 sec/min) = 57,600 J
(Notice that the units work out to give Watts * seconds, or Joules)

21b) Find energy in kWh (kilo-Watt hours)

W (kWh) = (P * t)/1000 ... (where t is time in hours)
W = (2W * 8 h)/1000 = 0.016 kWh

4.23 P = VI = W/t
t = W / (V*I) = 12 J / (3V * 2A) = 2 sec.

4.26 P = V**2 / R = (9mV)**2/(3 ohm) = 0.000027W = 27 micro W
(Notice: V**2 means 'V squared')

4.27 P = (I**2)*R ; I = sqrt (P/R) = sqrt (64W/4 ohm) = 4 A

4.29 P = V**2 / R ; V = sqrt (P*R) = sqrt (42mW * 2.2k ohm) = 9.61 V

4.34 P = V**2 / R ; V = sqrt (P*R) = sqrt (100W * 20 k ohm) = 1414.2 V
V = RI; I = V/R = 1414.2V / 20k ohm = 70.7 mA
double check: P = V * I = 1414.2V * 70.7mA = 100 W - OK

4.39 eff. = Po/Pi
.685 = 1.8hp/Pi ... Pi = 2.628hp * (746 W/ hp) = 1960.3 W
P = VI ... I = P/V = 1960.3W / 120 V = 16.34 A

4.54 P (kWh) = (P * t)/1000 (where t is in hours)
P = (30W * 3hr)/1000 = 0.09 kWh * 8 cents/kWh = 0.72 cents