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Problem set #1, EE 221, 08/27/2002 - /9/03/2002
Chapter 2, Problem 2.
Convert the following to engineering notation:
(a) 1,000,000 W (b) 12.35 mm
(c) 47,000 W (d) 0.00546 A
(e) 0.033 mJ ( f )5.33 ×10-6 mW
(g) 0.000000001 s (h) 5555 kW
(i) 32,000,000,000 pm
Chapter 2, Solution 2.
(a) 1 MW
(b) 12.35 mm
(c) 47. kW
(d) 5.46 mA
(e) 33 µJ
(f) 5.33 nW
(g) 1 ns
(h) 5.555 MW
(i) 32 mm
Chapter 2, Problem 9.
The waveform shown in Fig. 2.28 has a period of 10 s. (a) What is the average value of the current over one period?
(b) How much charge is transferred in the interval 1 < t < 12s? (c) If q (0) =0 sketch q(t),0 < t < 16 s.
Chapter 2, Solution 9.
Referring to Fig. 2.28,
(a) The average current over one period (10 s) is
iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10 =
800 mA
(b) The total charge transferred over the interval 0 < t < 12 s is (1 < t < 12) s
( qtotal
(c)
12
∫ i(t )dt = -4(2) + 2(2) + 6(2) + 0(4) –4(2)
= ∫ i (t )dt = -4(1) + 2(2) + 6(2) + 0(4) –4(2)
qtotal =
0
12
1
q (C)
=
=
0C
4C)
16
8
2
-8
-16
4
10
6
8
12
14
t(s)
16
Problem set #1, EE 221, 08/27/2002 - /9/03/2002
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Chapter 2, Problem 10.
Determine the power being absorbed by each of the circuit elements shown in
Fig. 2.29.
Chapter 2, Solution 10.
(a) Pabs = (+3.2 V)(-2 mA) = -6.4 mW
(b) Pabs = (+6 V)(-20 A) = -120 W
(or +6.4 mW supplied)
(or +120 W supplied)
(c) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60 W
(d) Pabs = (4 sin 1000t V)(-8 cos 1000t mA)
|
t = 2 ms
= +12.11 mW
Chapter 2, Problem 13.
The current-voltage characteristic of a silicon solar cell exposed to direct sun-light at twelve noon in Florida during
midsummer is given in Fig. 2.31. It is obtained by placing different size resistors across the two terminals of the
device, and measuring the resulting currents and voltages.
(a) What is the value of the short-circuit current?
(b) What is the value of the voltage at open circuit?
(c) Estimate the maximum power that can be obtained from the device.
Chapter 2, Solution 13.
(a) The short-circuit current is the value of the current at V = 0.
Reading from the graph, this corresponds to approximately 3.0 A.
(b) The open-circuit voltage is the value of the voltage at I = 0.
Reading from the graph, this corresponds to roughly 0.4875 V, estimating the curve as hitting the
x-axis 1 mm behind the 0.5 V mark. (note: 0.5 V is good enough)
(c) We see that the maximum current corresponds to zero voltage, and likewise, the maximum voltage
occurs at zero current. The maximum power point, therefore, occurs somewhere between these two points.
By trial and error,
Pmax is roughly (375 mV)(2.5 A) = 938 mW, or just under 1 W.
Problem set #1, EE 221, 08/27/2002 - /9/03/2002
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Chapter 2, Problem 15.
In the simple circuit shown in Fig. 2.33, the same current flows through each element. If Vx=1 V and VR =9 V,
compute: (a) the power absorbed by element A (b) the power supplied by each of the two sources (c) Does the total
power supplied = the total power absorbed? Is your finding reasonable? Why (or why not)?
Chapter 2, Solution 15.
We are told that Vx = 1 V, and from Fig. 2.33 we see that the current flowing through the dependent source
(and hence through each element of the circuit) is 5Vx = 5 A. We will compute absorbed power by using
the current flowing into the positive reference terminal of the appropriate voltage (passive sign convention),
and we will compute supplied power by using the current flowing out of the positive reference terminal of
the appropriate voltage.
(a) The power absorbed by element “A” = (9 V)(5 A) = 45 W
(b) The power supplied by the 1-V source = (1 V)(5 A) = 5 W, and
the power supplied by the dependent source = (8 V)(5 A) = 40 W
(c) The sum of the supplied power = 5 + 40 = 45 W
The sum of the absorbed power is 45 W, so
yes, the sum of the power supplied = the sum of the power absorbed, as we expect from the
principle of conservation of energy.
Chapter 2, Problem 17.
A simple circuit is formed using a 12-V lead-acid battery and an automobile headlight. If the battery delivers a total
energy of 460.8 watt-hours over an8-hour discharge period,(a) how much power is delivered to the headlight? (b)
what is the current flowing through the bulb? (Assume the battery voltage remains constant while discharging.)
Chapter 2, Solution 17.
The battery delivers an energy of 460.8 W-hr over a period of 8 hrs.
a) The power delivered to the headlight is therefore (460.8 W-hr)/(8 hr) = 57.6 W
b) The current through the headlight is equal to the power it absorbs from the battery divided by the voltage
at which the power is supplied, or
I = (57.6 W)/(12 V) = 4.8 A
Problem set #1, EE 221, 08/27/2002 - /9/03/2002
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Chapter 2, Problem 25.
It is not uncommon to see a variety of subscripts on voltages, currents, and resistors in circuit diagrams. In the
circuit in Fig. 2.38, the voltage vπ appears across the resistor named rπ . Compute voutif vs= 0.01 cos 1000t V.
Chapter 2, Solution 25.
The voltage vout is given by
vout
= -10-3 vp (1000)
= - vp
Since vp = vs = 0.01 cos 1000t V, we find that
vout
= - vp = -0.01 cos 1000t
V
Chapter 2, Problem 26.
A length of 18 AWG solid copper wire is run along the side of a road to connect a sensor to a central computer
system. If the wire is known to have a resistance of 53 , what is the total length of the wire? (Assume the
temperature is ~20C.)
Chapter 2, Solution 26.
18 AWG wire has a resistance of 6.39 Ohm / 1000 ft. Thus, we require 1000 (53) / 6.39 = 8294 ft of wire.
(Or 1.57 miles. Or, 2.53 km).