Hello,
I am not sure if I'm doing it right.
I get the solution for current as:
I(t) = I + (Io-I) * cos(t/sqrt(L*C))
Assuming the result is correct,
peak current for I(t) will depend on value of Io.
If Io > I, then Ipeak = Io, @ cos(t/sqrt(L*C)) = 1
If Io < I, then Ipeak = I, @ cos(t/sqrt(L*C)) = 0
Thanks,
Tanto
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I(t) solution can be expressed as:
I(t)=I+C1*sin(t/sqrt(L*C))+C2*cos(t/sqrt(L*C))
where C1 and C2 are constants obtained from:
I(0)=Io and dI(0)/dt=(V-Vo)/L
or
I(t)=I+C3*sin(t/sqrt(L*C)+C4)
where C3 and C4 are constants obtained from:
I(0)=Io and dI(0)/dt=(V-Vo)/L
Thanks Vojkan for your response.
I've another question in regard to the same problem. Can we assume that Io < I and Vo < V ?
Since the problem doesn’t give any relations
between Io and I, if we want to maximize (Io-I),
the safest thing would be to use an absolute value:
max(Io-I)=abs(Io-I)
min(Io-I)=-abs(Io-I)
which is true for Io< I, Io>I or I=Io
The same applies to Vo and V:
max(Vo-V)=abs(Vo-V)
min(Vo-V)=-abs(Vo-V)
Could anyone provide a hint for part d of this problem (once (d) is figured out part e is easy)?
I'm trying to find an easier way rather than the calculus approach of taking the derivative, finding a time, and plugging it into v(t) to find the value when I is maximized.
part d:
I is maximized when di/dt=0;
inductor voltage at that time is
v(L)=L*di/dt=0;
part e:
V is maximized when dv/dt=0;
capacitor current at that time is
i(C)=C*dv/dt=0;
I don't understand where the sine terms come from. I have i(t)=I+C*dv/dt and v(t)=V-L*di/dt. Also is the current source the load? Any hints?
Justin,
U are right but that is not in a compete time domain expression. You have to solve the small signal using the reactive elements and superimpose the current and voltage you get to the DC constats.
Rgrds,
Efrem
By eliminating v(t) from the two equations, 2nd order differential equation is obtained for i(t). Solution of the equation is the sine term (with amplutude and phase obtained from initial conditions).
The same applies to v(t): eliminating i(t), 2nd order diff equation is obtained for v(t). Solution is the sine term...
Brandon_Grainger,
Part d) and e): I first converted the A*sin(x)+B*cos(x)expression into sqrt(A^2+B^2)*sin(x+acos(A/sqrt(A^2+B^2)). This gives me a single sin function with a phase angle. Equate the angle, i.e (x+acos(A/sqrt(A^2+B^2), to pi/90-deg and zero to obtain max and min respectively. It could be pretty involving. In the case of the calculus, you have to further differentiate to decide if it is a max or a min. This is how I am pursuing. Have an idea?
I think I have a. d. and e. parts of this problem. But for b. and c. isnt it that the v' is the max if v" is -ve and i' is max if i" is -ve? I get these but what do you do with that? What is meant by find an "expression" for max v and max i?
Please forgive my lack of recall with respect to DE's. It's been a few years.
For parts b and c I understand that we take the derivative and set it equal to zero to find a max but I'm not sure what we solve for at that point. I assume it's t. I think that would give us the time at which v is at max. Am I correct in this approach?
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