CHAPTER 1ExercisesE1.1Charge = Current ( Time = (2 A) ( (10 s) = 20 CE1.2E1.4Energy = Charge ( Voltage = (2 C) ( (20 V) = 40 JBecause vab is positive, the positive terminal is a and the negative terminal is b. Thus the charge moves from the negative terminal to the positive terminal, and energy is removed from the circuit element.E1.5iab enters terminal a. Furthermore, vab is positive at terminal a. Thus the current enters the positive reference, and we have the passive reference configuration.E1.6(a)
(b) Notice that the references are opposite to the passive sign convention. Thus we have:
E1.7(a) Sum of currents leaving = Sum of currents enteringia = 1 + 3 = 4 AE1.8Elements A and B are in series. Also, elements E, F, and G are in series.ProblemsP1.1Broadly, the two objectives of electrical systems are:P1.2Four reasons that non-electrical engineering majors need to learn the fundamentals of EE are:P1.3Eight subdivisions of EE are:P1.4Responses to this question are varied.P1.5(a) Electrical current is the time rate of flow of net charge through a conductor or circuit element. Its units are amperes, which are equivalent to coulombs per second.P1.6(a) A conductor is anagolous to a frictionless pipe.P1.7P1.8*The reference direction for
points from a to b. Because
has a negative value, the current is equivalent to positive charge moving opposite to the reference direction. Finally since electrons have negative charge, they are moving in the referenceP1.12(a) The sine function completes one cycle for each
radian increase in the angle. Because the angle is
one cycle is completed for each time interval of 0.01 s. The sketch is:P1.13*P1.15The number of electrons passing through a cross section of the wire per second isP1.16*The charge flowing through the battery is
P1.17
P1.18P1.19*P1.20If the current is referenced to flow into the positive reference for the voltage, we say that we have the passive reference configuration. Using double subscript notation, if the order of the subscripts are the same for the current and voltage, weP1.21*(a) P
-vaia = 30 W Energy is being absorbed by the element.P1.22The amount of energy is
Because the reference polarity is positive at terminal a and the voltage value is negative, terminal b is actually the positive terminal. Because the charge moves from the negative terminal to the positive terminal, energP1.23*
.To increase the chemical energy stored in the battery, positive charge should move from the positive terminal to the negative terminal, in other words from a to b. Electrons move from b to a.P1.24P1.25P1.26*P1.27(a) P
60 W delivered to element A.P1.28*(a)P1.29The power that can be delivered by the cell is
In 71 hours, the energy is
Thus, the unit cost of the energy isP1.30The current supplied to the electronics is
The ampere-hour rating of the battery is the operating time to discharge the battery multiplied by the current. Thus, the operating time is
The energy delivered by the battery is
Neglecting the costP1.31A node is a point that joins two or more circuit elements. All points joined by ideal conductors are electrically equivalent. Thus, there are four nodes in the circuit at hand:
P1.32The sum of the currents entering a node equals the sum of the currents leaving.P1.33The currents in series-connected elements are equal.P1.34For a proper fluid analogy to electric circuits, the fluid must be incompressible. Otherwise the fluid flow rate out of an element could be more or less than the inward flow. Similarly the pipes must be inelastic so the flow rate is the same at alP1.35*Elements A and B are in series. Also, elements E and F are in series.P1.36(a) Elements C and D are in series.P1.37*At the node joining elements A and B, we have
Thus,
For the node at the top end of element C, we have
. Thus,
. Finally, at the top right-hand corner node, we have
Thus,
. Elements A and B are in series.P1.38*P1.39
Applying KCL, we findP1.40If one travels around a closed path adding the voltages for which one enters the positive reference and subtracting the voltages for which one enters the negative reference, the total is zero.P1.41(a) Elements A and B are in parallel.P1.42*Summing voltages for the lower left-hand loop, we have
which yields
Then for the top-most loop, we have
which yields
Finally, writing KCL around the outside loop, we have
which yieldsP1.43We are given
Applying KVL, we findP1.44*Applying KCL and KVL, we haveP1.45(a) In Figure P1.28, elements C, D, and E are in parallel.P1.47(a) The voltage between any two points of an ideal conductor is zero regardless of the current flowing.P1.48Four types of controlled sources and the units for their gain constants are:P1.49Provided that the current reference points into the positive voltage reference, the voltage across a resistance equals the current through the resistance times the resistance. On the other hand, if the current reference points into the negative voltP1.50*P1.51P1.52The resistance of the copper wire is given by
, and the resistance of the nichrome wire is
. Taking the ratios of the respective sides of these equations yields
. Solving for
and substituting values, we have
P1.53P1.54P1.55*P1.56The power delivered to the resistor isP1.57The power delivered to the resistor isP1.58Equation 1.10 gives the resistance asP1.59(a) The voltage across the voltage source is 10 V independent of the current. Thus, we have v ( 10 which plots as a vertical line in the v(i plane.(b) The current source has i ( 2 independent of v, which plots as a horizontal line in the v(i plane.P1.60*(a) Not contradictory.P1.61The power for each element is 20 W. The current source delivers power and the voltage source absorbs it.P1.62*P1.63This is a parallel circuit and the voltage across each element is 10 V positive at the top end. Thus, the current through the resistor isP1.64*P1.66(a) The 3-( resistance, the 2-( resistance, and the voltage source Vx are in series.P1.67First, we haveP1.68(a) No elements are in series.P1.69P1.70*(a) Applying KVL, we have
, which yieldsP1.71
V
AP1.72Consider the series combination shown below on the left. Because the current for series elements must be the same and the current for the current source is 2 A by definition, the current flowing from a to b is 2 A. Notice that the current is not afP1.73Consider the parallel combination shown below. Because the voltage for parallel elements must be the