请教两道物理电学题。200分。英文版的。求各位解答。。求详细。谢谢了.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. (a) Vc(0+)=Vc(0-)=30*(4+6)/(4+6+5)=20V.
(b) Equivalent capacitance C'=4+4=8µF. (The capacitor should be 4µF instead of 4F. 4F is too big a value for capacitance in reality).
Equivalent resistance R'=4+1/(1/6+1/6))=7Ω.
The time constant τ=R'*C'=56µs.
(c) Vc(∞)=4*1/(1/6+1/6)=12V.
So Vc(t)=Vc(∞)+(Vc(0+)-Vc(∞))e^(-t/τ)=12+8e^(-t/56e-6).
2. (a) When S2 is kept closed, VL=L*dI/dt=L*0=0V so there is no current flowing through the 50Ω resistor. So IL(0-)=2A. The inductor current cannot be changed instantly when S1, S2 are flipped so IL(0+)=IL(0-)=2A.
(b) Equivalent resistance R'=20+1/(1/20+1/20)=30Ω.
So the time constant τ=L/R'=1µs.
(c) IL(∞)=30/(20+1/(1/20+1/20))*20/(20+20)=0.5A.
So IL(t)=IL(∞)+(IL(0+)-IL(∞))e^(-t/τ)=0.5+1.5e^(-t/1e-6).
(d) IL(τ)=0.5+1.5e^(-1)=1.05A.
IL(2τ)=0.5+1.5e^(-2)=0.703A.
IL(3τ)=0.5+1.5e^(-3)=0.575A.
IL(5τ)=0.5+1.5e^(-5)=0.510A.
9.
Since VR1=Vs-V1, we just solve the set of equations:
KCL at node V1: (Vs-V1)/R1+(V2-V1)/R3+(V2+Av*(Vs-V1)-V1)/R2=0
KCL at node V2: (V1-V2)/R3-(V2+Av*(Vs-V1)-V1)/R2-V2/R4=0
, which equals
(-1/R1-1/R3-(Av+1)/R2)*V1+(1/R3+1/R2)*V2=-Vs/R1-Av*Vs/R2
(1/R3+(Av+1)/R2)*V1+(-1/R3-1/R2-1/R4)*V2=Av*Vs/R2
And we get V1=4.912V, V2=8.757mV.
So the answer VR4=V2=8.757mV.
10.
When only the current source is present, we should treat the voltage source as a line.
V1= IB*(1/RB+1/RG+1/R)^(-1)=1.2V.
When only the voltage source is present, we should treat the currentsource as an open loop.
V2=V*(RB*R)/(RB+R)/(RB*R/(RB+R)+RG)=6V.
So use the principle of superposition, if both sources are present, the answer is
V=V1+V2=7.2V.
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