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