Otis, explain to me how quantum mechanics work in DETAIL please.
There is some tiny stuff, then it does stupid stuff in a quantized fashion. That is, with discrete results for the thingamajiggers.
Also,
Example 7
Find the charge Q(t) and current I(t) in the circuit with E(t) = Eo with the switch closed at t = 0. Also, show
lim┬(t→∞)〖Q(t)=E_o C,lim┬(t→∞)〖I(t)〗 〗=0
Voltage loop equation
RQ^'+1/C Q=E_o
Transient charge and root
Rr+1/C=0,r=(-1)/RC
Q_tr=c_1 e^((-t)/RC)
Steady periodic charge
Q_sp=AE_o
〖Q^'〗_sp=0
1/C AE_o=E_o
A=C
Q_sp=E_o C
General solution
Q(t)=Q_tr+Q_sp=c_1 e^((-t)/RC)+E_o C
Particular solution given Q(0) = 0
Q(0)=0=c_1+E_o C,c_1=-E_o C
Q(t)=-E_o Ce^((-t)/RC)+E_o C
I(t)=Q^' (t)=E_o/R e^((-t)/RC)
Now that we have the functions for charge and current we can check the limits of each to see if they have the values we are interested in.
lim┬(t→∞)〖Q(t)=〗-E_o Ce^((-∞)/RC)+E_o C=0+E_o C=E_o C
The above limit shows that the transient charge in a RC circuit approaches zero in the same fashion that the transient current in an RL circuit does. The long term behavior of the function will resemble the steady periodic charge.
〖 lim┬(t→∞)〖I(t)〗 〗=E_o/R e^((-∞)/RC)=0
In basic circuit brown townysis it is taught that as t → +∞ a capacitor will act as an open circuit which means no current will flow. The above limits show that as t → +∞ the charge across the plates of a capacitor will approach a constant value. If the charge is a constant value then its derivative is zero. The derivative of the charge function is also the current function. Our derivation in this example shows that the assumption about a capacitor’s behavior as t → +∞ is correct.
Example 8
R = 10, C =.02, Q(0) = 0, E(t) = 100e-5t. Find Q(t) and I(t). Find the maximum charge and the time at which it occurs.
Voltage loop equation
10Q^'+50Q=100e^(-5t)
Characteristic equation and root
10r+50=0,r=-5
Transient charge
Q_tr=c_1 e^(-5t)
Steady periodic charge
Q_sp=Ae^(-5t)
There is duplication of our transient and steady periodic guess, so we will multiply the periodic function by t to resolve this.
Q_sp=A〖te〗^(-5t)
〖Q'〗_sp=Ae^(-5t)-5Ate^(-5t)
10〖Q^'〗_sp+50Q_sp=100e^(-5t)
10Ae^(-5t)-50Ate^(-5t)+50A〖te〗^(-5t)=100e^(-5t)
A=10
General solution
Q(t)=c_1 e^(-5t)+10te^(-5t)
Particular solution for Q(0) = 0
Q(0)=0=c_1
Q(t)=10te^(-5t)
I(t)=Q^' (t)=10e^(-5t)-50te^(-5t)
To find the maximum charge we check at what points I(t) = 0. As with Example 4 it is seen that t = ∞ is an equilibrium solution.
10e^(-5t)-50te^(-5t)=0
10e^(-5t)=50te^(-5t)
t=1/5 s
Q_max=Q(1⁄5)=10(1/5) e^(-5(1⁄5) )=2e^(-1)
Example 9
R = 200, C = 2.5 x 10-4, Q(0) = 0, E(t) = 100cos(120t). Find Q(t) and I(t). Find the amplitude of the steady state current.
Voltage loop equation
200Q^'+4000Q=100cos〖(120t)〗
Characteristic equation and roots
200r+4000=0,r=-20
Transient charge
Q_tr=c_1 e^(-20t)
Steady periodic charge
Q_sp=A cos〖(120t)+B sin〖(120t)〗 〗
〖Q'〗_sp=-120A sin〖(120t)〗〖+120B cos(120t) 〗
200〖Q'〗_sp+4000Q_sp=100cos〖(120t)〗
-24000A sin〖(120t)〗〖+24000B cos〖(120t)+4000A cos〖(120t)+4000B sin〖(120t)=100 cos〖(120t)〗 〗 〗 〗 〗
-24000A+4000B=0
4000A+24000B=100
A=1/1480 B=6/1480
Q_sp=1/1480 cos〖(120t)+〖6/1480 sin〗〖(120t)〗 〗
General solution
Q(t)=c_1 e^(-20t)+1/1480 cos〖(120t)+〖6/1480 sin〗〖(120t)〗 〗
Particular solution
Q(0)=0=c_1+1/1480, c_1=-1/1480
Q(t)=-1/1480 e^(-20t)+1/1480 cos〖(120t)+〖6/1480 sin〗〖(120t)〗 〗
I(t)=Q^' (t)=1/74 e^(-20t)+18/37 cos〖(120t)+〗 (-3)/37 sin〖(120t)〗
From the above function I(t) we can pick out the steady periodic part and determine its amplitude using the trigonometric properties we used in Example 6.
I_sp (t)=18/37 cos〖(120t)+〗 (-3)/37 sin〖(120t)〗
The phase α can be determined by
α=tan^(-1)〖(B/A)=tan^(-1)((-3)/18)=-0.165〗
B is negative and A is positive so α is in the fourth quadrant, so
α=2π-0.165=6.118 radians
The magnitude of the amplitude of Isp(t) is
C/37=√(A^2 〖+B〗^2 )=√(〖18〗^2+〖(-3)〗^2 )=(3√37)/37=3/√37
I_sp=3/√37 cos(120t-6.118)
Example 10
E(t) = E_o cos(ωt) at t = 0 when switch is closed. Qsp(t) = Acos(ωt) + Bsin(ωt). Q(0) = 0. Show that the steady periodic charge is
Q_sp (t)=(E_o C)/√(1+ω^2 R^2 C^2 ) cos(ωt-β),β=tan^(-1)(ωRC)
Voltage loop equation
RQ^'+(1/C)Q=E(t)
Steady periodic charge
Q_sp (t)=A cos〖(ωt)+B sin(ωt) 〗
〖Q^'〗_sp (t)=-ωA sin〖(ωt)+ωB cos(ωt) 〗
-ωRA sin〖(ωt)+ωRB cos〖(ωt)+A/C cos〖(ωt)+B/C sin(ωt) 〗 〗 〗=E_o
A/C+RωB=E_o
-RωA+B/C=0
A=E_o/C[(R^2 ω^2 )+(1/C^2 ) ] =E_o/[(CR^2 ω^2 )+(1/C) ]
B=(E_o Rω)/(R^2 ω^2+1/C^2 )
Q_sp (t)=E_o/[(CR^2 ω^2 )+(1/C) ] cos〖(ωt)+(E_o Rω)/(R^2 ω^2+1/C^2 ) sin(ωt) 〗
Q_sp (t)=(E_o C)/[(C^2 R^2 ω^2 )+1] cos〖(ωt)+(E_o C^2 Rω)/(C^2 R^2 ω^2+1) sin(ωt) 〗
Q_sp (t)=((E_o C)/(C^2 R^2 ω^2+1))[cos〖(ωt)+RCω sin(ωt) 〗 ]
As with previous problems we can combine both trig functions into a single term. First, the phase angle β
β=tan^(-1)〖(B/A)=tan^(-1)(ωRC/1) 〗
The magnitude of coefficient D of the single term
D=√(1^2+(ωRC)^2 )
Q_sp (t)=((E_o C)/(C^2 R^2 ω^2+1)) √(1^2+(ωRC)^2 ) [cos(ωt-β) ]=(E_o C)/√(1+ω^2 R^2 C^2 ) cos(ωt-β)
β=tan^(-1)(ωRC)
