200分.英文版的..ans是答案

200分.英文版的..ans是答案
200分.英文版的..ans是答案

200分.英文版的..ans是答案
【Question 4】
Solution :

Effective Resistance of the 6-Ω resistor and two parallel connected 4-Ω resistor is
R₁= 6 Ω + 2 Ω = 8 Ω
Effective Resistance of R₁and the 24-Ω resistor in parallel connection is
R₂= 8×24/(8+24) = 6 Ω
Effective Resistance of R₂and the third 4-Ω resistor in series connection is
R₃= 4 Ω + 6 Ω = 10 Ω
Effective Resistance of R₃and the 15-Ω resistor in parallel connection is
R₄= 15×10/(15+10) = 6 Ω
The total effective Resistance of R₄and the R-Ω resistor in series connection is
Rt = (6+R) Ω
Total current is
I₁= V/Rt = 25/(6+R) [1]

The power absorbed by the 15-Ω resistor is 15 W,
so, the PD cross the 15-Ω resistor is
V₁= sqrt(P×R) = sqrt(15×15) = 15 V
So, the PD cross the R-Ω resistor is
V₂= 25 - 15 = 10 V
The current carried by the R-Ω resistor is
I₂= 10/R [2]

Let I₁= I₂, we have
25/(6+R) = 10/R, 5R = 12 + 2R
R = 4 (Ω)


【Question 5】
Solution :
(a)
i₁+ i₂= - 3 (A)
i₂R₃- i₁R₂= 12
(-3 - i₁)R₃- i₁R₂= 12
i₁= (-12 - 3R₃)/(R₂+ R₃) = (-12 - 15)/15 = -1.8 (A)
i₂= - 3 - i₁= - 3 + 1.8 = -1.2 (A)

(b)
The PD cross R₄= 12 V
The PD cross R₃= 1.2×5= 6 V
The PD cross R₃and R₄, also PD cross R₂= 6 + 12 = 18 V
The PD cross R₁= 3×25 = 75 V
Total out-put PD for 3A current source = 75 + 18 = 93 V
The power delieved by 3A current source
= IV = 3×V = 3×93 = 297 (W)
The power supplied by 12V voltage source
= IV = (-1.2 + 12/7)×12 = 6.171 (W)

(c)
Total power dissipated by the sircuit is 279 + 6.171 = 285.171 (W)


【Question 6】
Solution :
(a)
Effective resistance of 7-Ω and 5-Ω resistors is 12 Ω.
i = V/R = 24/12 = 2 (A)
The current suppulied by the dependent current source is 0.5×2² = 2 (A)
Total external resistance = 15 + 7 + 5 = 27 (Ω)
Power delieved by the dependent current source is
P = I²R = 2²×27 = 108 (W)

(b)
Power delieved by the voltage source is
P = - IV + i²×(7+5) = - 2×24 + 4×12 = - 48 + 48 = 0

1. 6欧下面两个4欧并联为2欧，再与6欧串联为8欧。8欧与24欧并联为8*24/(8+24)=6欧。6欧再与4欧串联为10欧。
已知通过15欧电阻消耗功率为15W，由U^2/R=P可知，U=15V。可知10欧的通过电流为15/10=1.5A。于是通过R的电流为1+1.5=2.5A。由KVL可知，加在R的电压为25-15=10V。
(25-15)/2.5＝4欧
2. 对R2...

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1. 6欧下面两个4欧并联为2欧，再与6欧串联为8欧。8欧与24欧并联为8*24/(8+24)=6欧。6欧再与4欧串联为10欧。
已知通过15欧电阻消耗功率为15W，由U^2/R=P可知，U=15V。可知10欧的通过电流为15/10=1.5A。于是通过R的电流为1+1.5=2.5A。由KVL可知，加在R的电压为25-15=10V。
(25-15)/2.5＝4欧
2. 对R2，R3和电压源列KVL，顺时针并假定电阻正极都在箭头指向的反方向，有
12-v2+v3=0
其中v2=10*I1，v3=5*I2
考虑到3+i1+i2=0，联立上面两式求得i1=-1.8，i2=-1.2

求输出功率要对电源用P=IU即可
对电流源，R1和R2列KVL有V+3*25+10*1.8=0，可得V=93V，于是P＝93*3=279W
电压源除了i2还有电流i3供给R4，i3=12/7=1.714A。于是P=12*(-1.2+1.713)=6.17W
总消耗的能量就是两个电源消耗能量的和279+6.17=285.17W
此题一个很关键的地方在于电流的方向和电源的正负。记住电阻正极都在箭头指向的反方向，这个时候可能会得到负的电压（在电流为负的时候），切记要保留符号计算。
3. 首先求i，i=24/(7+5)=2A，因此电流源为0.5*4=2A
根据KVL，电流源的电压为V=2*15+24=54V，于是功率为54*2=108W
根据KCL，电流源没有输出任何电流，所以功率为0

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