prove L.(di/dt) + R.i = V.cos(wt)

[markbrock30]

Given v(t) = V.cos(wt).

And a circuit consisting of v(t), and an R and L component.

How can I prove that the differential equation for the current i(t) is:

L.(di/dt) + R.i = V.cos(wt)


Cheers,

Mark.

 

[stapel]

Given v(t) = V.cos(wt), [a]nd a circuit consisting of v(t), and an R and L component. How can I prove that the differential equation for the current i(t) is L.(di/dt) + R.i = V.cos(wt)[?]

What do the decimal points mean? For instance, what is the definition of "V point cosine of wt"?

What is "an R and L component", and what assumptions, definitions, etc, are required in order to incorporate this information?

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you!

Eliz.

 

[Deleted member 4993]

Given v(t) = V.cos(wt), and a circuit consisting of v(t), and an R and L component.

How can I prove that the differential equation for the current i(t) is:

L.(di/dt) + R.i = V.cos(wt) <<<< This is derived from laws of physics where L.di/dt is the inductive potential drop and R.i is the resistive potential drop.

 

[markbrock30]

What do the decimal points mean?

The decimal points are multiples; L*(di/dt) + R*i = V*cos(w*t) may be clearer.

what is the definition of "V point cosine of wt"?

This problem is taken from an electrical circuit. The voltage (v(t)) is decribed at any point in time by V*cos(w*t)

What is "an R and L component"

R is a resistor, and L is an inductor. (both with unknown values).

What are your thoughts? What have you tried?

I have an example that I have been trying to work from with a capacitor instead of an inductor. I know that Kirchoff's law states that the sum of the voltages across R and L must equal v(t).

Therefore... if R*i + (1/C).i = v(t), then I figured that R.i + L.i = v(t)???

Where R is the resistor value, C is the capacitor value, i is the current, v(t) is the voltage in respect to time.

In the example, they differentiated the equation above to get the differential equation for i.

So the circuit has a voltage source (v(t)), an inductor (L) and a resistor (R) - I am trying to prove that the the differential equation for the current (i(t)) is:
L.(di/dt) + R.i = V.cos(wt).

How far have you gotten? Where are you stuck?

Not far, and I'm stuck right at the start. I don't think this is a long question, I just don't undestand the process.

Cheers,
Mark

 

[Deleted member 4993]

I have an example that I have been trying to work from with a capacitor instead of an inductor. I know that Kirchoff's law states that the sum of the voltages across R and L must equal v(t).

Therefore... if R*i + (1/C).i = v(t), <<< That is not correct


then I figured that R.i + L.i = v(t)??? <<< That is not correct

This is a physics problem not a mathematics problem.

The voltage drop across a resistor (R) is proportional to the amount of current (i) - this is Ohm's law

The voltage drop across an inductor(L) is proportional to the rate of change of current (di/dt) - Ampere's(?) law

This should have been covered in your physics class - before you tackle this type of problem. The next level of complexity will come in you introduce a third component in the circuit - a capacitor (C). That follows a different law. Then you put those in parallel or series - more fun to follow. But learn the physics behind it - before you get bogged down with mathematics.

 

[markbrock30]

Solve L*(di/dt) + R*i = V*cos(w*t) to find i(t)

Problem:

Let w = 2000
V = 6V
L = 0.2H
R = 400

Given the differential equation L*(di/dt) + R*i = V*cos(w*t)

Find i(t). - The above equation is true for t > 0

This is what I have come up with. - NOTE: When im integrating ive used the '!' symbol

-> 0.2*(di/dt) + 400*i = 6*cos(w*t) divide by 0.2
-> (di/dt) + 2000*i = 30*cos(w*t)

I solved the above using the steady state and transient method.

For the transient part:

-> di/dt + 2000*i = 0

-> di = -2000*i*dt

-> !(1/i)*di = !-2000*dt '!' means integrate

-> ln(i) = -2000t + c

-> i(transient) = e^-2000t + e^c

-> i(transient) = A*e^-2000t

For the steady state part:

Ohms law states that the current i = V/Z, where V is voltage, and Z is impedance.

So, V = 6<0 V The '<' is the polar form (or phasor some might know it as) representation of the complex angle associated with the voltage, so 6V, 0 angle

Z = R + j*w*L
Z = 400 + j*2000*0.2
Z = 400 + j400
Z = 565.685<0.785 angle is in radians

-> i = V/Z

-> i = 6<0 / 565.685<0.785

-> i = 10.61<-0.785 mA

-> i = 10.61*sin(w*t - 0.785) mA

Therefore I can derrive i(t):

-> i(t) = i(steady state) + i(transient)

-> i(t) = A*e^-2000t + 10.61*sin(w*t - 0.785) mA

To find A make t = 0

-> i(0) = A(1) + (10.61*10^-3)*sin(-0.785)

-> 0 = A - 7.5*10^-3

-> A = 7.5*10^-3

So the final equation for i(t) is:

-7.5*e^-2000t + 10.61*sin(w*t - 0.785) mA

However this is not working out - I should be able to substitute this equation back into the original:

-> L*(di/dt) + R*i = V*cos(w*t)

L = 0.2
R = 400
i(t) = 7.5*e^-2000t + 10.61*sin(2000*t - 0.785) mA
i'(t) = -15*e^-2000t + 21.22*cos(2000*t - 0.785)
V = 6

-> 0.2*(-15*e^-2000t + 21.22*cos(2000*t - 0.785)) + 400((7.5*10^-3)*e^-2000t + (10.61*10^-3)*sin(2000*t - 0.785)) = 6*cos(2000*t)

Let t = 1

-> 0.2(0 + 8.434) + 400(0 + (9.8677*10^-3)) = 6cos(2000)

-> 1.6869 + 3.947 = -2.205

This does not satisfy the equation!!!

Can anyone see where I went wrong?


Cheers,

Mark.

 

[markbrock30]

I have an example that I have been trying to work from with a capacitor instead of an inductor. I know that Kirchoff's law states that the sum of the voltages across R and L must equal v(t).

Therefore... if R*i + (1/C).i = v(t), <<< That is not correct


then I figured that R.i + L.i = v(t)??? <<< That is not correct

This is a physics problem not a mathematics problem.

I can assure you this goes on to be a nasty mathematics problem .

I talked to my math teacher today and found that the statement I made above is the correct answer to the question (worth 1 mark too... ). The equations though as you rightly stated are wrong - Just a product of my over-complication of any equation that's thrown in front of me...

The second part of the equation is a bit trickier, I have posted the Q, and my working so far above this message.

I'd appreciate if someone could let me know where i've gone wrong.


cheers,

Mark.

 

[Deleted member 4993]

If you are not worried about derivation - then it is a relatively simple Ordinary Differential Equation(ODE).

you have

\(\displaystyle \frac {di}{dt}\, +\, w\cdot i = 30\cdot cos(wt)\)

This is a first order differential equation, which transforms intoafter multiplying by integrating factor)

\(\displaystyle \frac {d}{dt}(e^{wt}\cdot i)\, = 30e^{wt}\cdot cos(wt)\)

Integrating both sides

\(\displaystyle e^{wt}\cdot i\,\, = \frac{15}{w}\cdot e^{wt}\cdot [cos(wt) + sin(wt)] + C_1\)

\(\displaystyle i\,\, = \frac{15}{w}\cdot [cos(wt) + sin(wt)] + C\)

at t=0, i=0

\(\displaystyle i\,\, = \frac{15}{w}\cdot [1- [cos(wt) + sin(wt)]]\)

\(\displaystyle i\,\, = \frac{15}{w}\cdot [1- \sqrt{2} \cdot sin(\frac{\pi}{4} \, + \, wt) ]\)

 

[markbrock30]

I can't even begin to imagine how you think!!!

I ended up with:

i(t) = -7.5*10^-3*exp(-2000*t) + 10.61*10^-3*sin(2000*t + 0.785)

which is the right answer...

but I you got the right answer as well in 1/4 the steps I took .

Thanks anyway.