电路英文版第二章答案

to the bottom terminal in the left hand branch of 50 − (−l50) = 200V. But the voltage drop between these same terminals in
the right hand branch is 250 V, due to the voltage source in that branch. Therefore, the interconnection is invalid.
P18V = −(18)(5 × 10−3) = −0.09W (del) P7V = (7)(5 × 10−3) = 0.035W (abs) P15A = (11)(5 × 10−3) = 0.055W (abs)
Pabs = Pdel = 0.09W P2.3 The interconnection is valid.
[b] NO, because the voltage drop between the top terminal and the bottom terminal cannot be determined. For example, define v1, v2, and v3 as shown:
1
The voltage drop across the left branch, the center branch, and the right branch must be the same, since these branches are connected at the same two terminals. This requires that
[b] ∆v = 20V; ∆i = 10mA;
3
[c]
R
=
∆v ∆i
=
2kΩ
2i1 = 3is, i1 = 1.5is 40 = i1 + is = 2.5is, is = 16mA
[d]vs (open circuit) = (40 × 10−3)(2 × 103) = 80V [e| The open circuit voltage can be found in the table of values (or from the plot) as the value of the voltage vs when the current is = 0. Thus,
=
4500 32
=
8000 42
=
12500 52
=
18000 62
=
500Ω
P2.13
Vbb = no-load voltage of battery Rbb = internal resistance of battery Rx = resistance of wire between battery and switch Ry = resistance of wire between switch and lamp A Ra = resistance of lamp A Rb = resistance of lamp B Rw = resistance of wire between lamp A and lamp B Rg1 = resistance of frame between battery and lamp A Rg2 = resistance of frame between lamp A and lamp B S = switch
P2.8 The interconnection is invalid. In the middle branch, the value of the current i∆ must be −25A, since the 25A current source supplies current in this branch in the direction opposite the direction of the current i∆. Therefore, the voltage supplied by the dependent voltage source in the left hand branch is 6(−25) ± −150V. This gives a voltage drop from the top terminal
P2.14[a] Plot the v − i characteristic:
2
140
120
100
80
vt (V)
60
40
20
0
−20
−40
0
1
2
3
4
5
6
7
8
it (A)
From the plot:
R
=
∆v ∆i
=
(10 − (−30)) (2 − 0)
= 20Ω
When it = 0, vt = −30V; therefore the ideal current source must have a current of −30/20 = 1.5A.
Pabs = Pdel = 0.125W [d| The interconnection is valid, but in this circuit the voltage drop across the 15mA current source is 11V, positive at the left; 7V source is absorbing, the 18V source is delivering, and the 15mA source is absorbing.
Hence any combination of v1 and v2 such that v1 + v2 = 50V is a valid solution. P2.6
The interconnection is invalid. The voltage drop between the top terminal and the bottom terminal on the left hand side is due to the 6 V and 8 V sources, giving a total voltage drop between these terminals of 14 V. But the voltage drop between the top terminal and the bottom terminal on the right hand side is due to the 4 V and 12 V sources, giving a total voltage drop between these two terminals of 16 V. The voltage drop between any two terminals in a valid circuit must be the same, so the interconnection is invalid.
[b] We attach a 5Ω resistor to the device model developed in part (a): Write a KCL equation at the top node: 1.5 + it = i1
Write a KVL equation for the right loop, in the direction of the two currents, using Ohm’s law: 20i1 + 5it = 0
Chapter 2 Circuit Elements P2.1[a] Yes, independent voltage sources can carry the 5mA current required by the connection; independent current source can support any voltage required by the connection, in this case 25V, positive at the right. [b] 18V source: absorbing
20 + v1 = v2 + 100 = v3
But this equation has three unknown voltages, so the individual voltages cannot be determined, and thus the power of the sources cannot be determined.
7V source: absorbing 5mA source: delivering [c]P18V = (18)(5 × 10−3) = 0.09W (abs) P7V = (7)(5 × 10−3) = 0.035W (abs) P15A = −(25)(5 × 10−3) = −0.125W (del)
P2.7[a] Yes, each of the voltage sources can carry the current required by the interconnection, and each of the current sources can carry the voltage drop required by the interconnection. (Note that i∆ = −8A.)
The total power developed by the voltage sources is P = −(100 − 40) × 5 = −300W , which means that the voltage sources are absorbing energy.
P2.5 The interconnection is valid, since the voltage sources can carry the 10 A current supplied by the current source, and the current sources can carry whatever voltage drop is required by the interconnection. In particular, note the the voltage drop across the three sources in the right hand branch must be the same as the voltage drop across the 20 A current source in the middle branch, since the middle and right hand branch are connected between the same two terminals. In particular, this means that (the voltage drop across the middle branch) v1 = 100V − 50V − v2(where v2 is the voltage drop across the right hand branch)
Combining the two equations and solving, 20(1.5 + it) + 5it = 0 sቤተ መጻሕፍቲ ባይዱ 25it = −30; thus it = −1.2A NOW calculate the power dissipated by the resistor:
p5Ω = 5i2t = 5(−1.2)2 = 7.2W P2.16[a]
P2.12 The resistor value is the ratio of the power to the square Using the values for power and current in Fig. P2.12(b),
of
the
current:
R
=
p i2
.
500 12
=
2000 22