Derive Inductor current in RL circuit w/ AC source: Vs.sin(wt) = L.(di/dt) + i.R.

SimsPhilly

New member
Joined
Sep 28, 2018
Messages
6
Hi all,

Here is my differential equation:

Vs.sin(wt) = L.(di/dt) + i.R.

With
- Vs.sin(wt) my voltage source
- L my inductor
- R my Resistor
all in series.

Question: derive the inductor current i.

Here is my solution: i(t) = (Vs/R).sin(wt) - (Vs/R).sin(wt).e^(-(R/L).t)

However, the solution is not final yet, and this where I'm stuck.
The solution should be in the following form:

i(t) = (Vs/Z).sin(wt - phi) + A.e^(-(R/L).t)

With
- Z = sqrt((wL)^2 + R2). by trigonometry
- phi = tan(-1)(wL/R)
- A to be defined using initial condition i(0) = 0 such as A=(Vs/Z).sin(phi)

I feel I'm close to it but can't find my way out of this.
Your help would be much appreciated!

Thank you

Sims
 
Here is how I would do this problem:

The differential equation is Ldidt+Ri=Vssin(ωt)\displaystyle L\frac{di}{dt}+ Ri= Vs sin(\omega t). The associated homogeneous equation is Ldidt+Ri=0\displaystyle L\frac{di}{dt}+ Ri= 0 or Ldii=dt\displaystyle L\frac{di}{i}= -dt. Integrating both sides, Lln(i)=t+C\displaystyle L ln(i)= -t+ C or i(t)=Cet/L\displaystyle i(t)= C'e^{-t/L}.

To find a solution to the entire equation, we look for a solution of the form i(t)=Acos(ωt)+Bsin(ωt)\displaystyle i(t)= A cos(\omega t)+ Bsin(\omega t) so that didt=Aωsin(ωt)+Bsin(ωt)\displaystyle \frac{di}{dt}= -A\omega sin(\omega t)+ B sin(\omega t) and the equation becomes ALωsin(ωt)+BLωcos(ωt)+RAcos(ωt)+RBsin(ωt)=(RBALω)sin(ωt)+(BLω+RA)cos(ωt)=Vssin(ωt)\displaystyle -AL\omega sin(\omega t)+ BL\omega cos(\omega t)+ RA cos(\omega t)+ RB sin(\omega t)= (RB- AL\omega)sin(\omega t)+ (BL\omega+ RA)cos(\omega t)= Vs sin(\omega t).
Your solutions don't look right to me but to answer your question, you need to use the trig identity cos(A+B)=sin(A)cos(B)+cos(A)sin(B)\displaystyle cos(A+ B)=sin(A)cos(B)+ cos(A)sin(B). Then also cos(AB)=sin(A)cos(B)cos(A)sin(B)\displaystyle cos(A- B)= sin(A)cos(B)- cos(A)sin(B) because sine is an "odd" function while cosine is an "even" function. Adding the two equations, 2sin(A)cos(B)=cos(A+B)+cos(AB)\displaystyle 2sin(A)cos(B)= cos(A+ B)+ cos(A- B).

Here is how I would do this problem:

Your equation is Ldidt+Ri=Vssin(ωt)\displaystyle L\frac{di}{dt}+ Ri= Vs sin(\omega t). The Associated homogeneous equation is didt=Ri\displaystyle \frac{di}{dt}= -Ri which can be written as dii=RLdt\displaystyle \frac{di}{i}= -\frac{R}{L}dt. Integrating both sides, ln(i(t))=RLt+C\displaystyle ln(i(t))= -\frac{R}{L}t+ C or i(t)=CeRLt\displaystyle i(t)= C'e^{-\frac{R}{L}t}.

To find a solution to the entire equation, look for a solution of the form i(t)=Asin(ωt)+Bcos(ωt)\displaystyle i(t)= A sin(\omega t)+ B cos(\omega t). Then didt=Aωcos(ωt)+Bωsin(ωt)\displaystyle \frac{di}{dt}= -A\omega cos(\omega t)+ B\omega sin(\omega t) and the equation becomes ALωcos(ωt)+BLωsin(ωt)+ARsin(ωt)+BRcos(ωt)=(BRAL)cos(ωt)+(BL+AR)sin(ωt)=Vssin(ωt)\displaystyle -AL\omega cos(\omega t)+ BL\omega sin(\omega t)+ AR sin(\omega t)+ BR cos(\omega t)= (BR- AL)cos(\omega t)+ (BL+ AR) sin(\omega t)= Vs sin(\omega t).

In order that this be true for all t we must have the coefficients of sin(ωt)\displaystyle sin(\omega t) and of cos(ωt)\displaystyle cos(\omega t) the same for both sides: RBALω=Vs\displaystyle RB- AL\omega = Vs and BL+RA=0\displaystyle BL+ RA= 0. To eliminate A, multiply the first equation by R, the second equation by Lω\displaystyle L\omega and add: R2BRAL=RVs\displaystyle R^2B- RAL= RVs added to BL2ω+RAL=0\displaystyle BL^2\omega+ RAL= 0 is (R2+L2ω)B=RVs\displaystyle (R^2+ L^2\omega)B= RVs so B=RVsR2+L2ω\displaystyle B= \frac{RVs}{R^2+ L^2\omega}. Then BL+RA=0\displaystyle BL+ RA= 0 becomes RA=BL=RLVsR2+L2ω\displaystyle RA= -BL= -\frac{RLVs}{R^2+ L^2\omega} and then A=LVsR2+L2ω\displaystyle A= -\frac{LVs}{R^2+ L^2\omega}.

So the general solution to the differential equation is i(t)=LVsR2+L2ωcos(ωt)+RVsR2+L2ωsin(ωt)+Cet/L\displaystyle i(t)= -\frac{LVs}{R^2+ L^2\omega}cos(\omega t)+ \frac{RVs}{R^2+ L^2\omega}sin(\omega t)+ C' e^{-t/L}.

If we also have the initial condition that i(0)= 0 then
i(o)=LVsR2+L2ω+C=0\displaystyle i(o)= -\frac{LVs}{R^2+ L^2\omega}+ C'= 0 so that C=LVsR2+L2ω\displaystyle C'= \frac{LVs}{R^2+ L^2\omega} and the solution to the initial value problem is
i(t)=LVsR2+L2ωcos(ωt)+RVsR2+L2ωsin(ωt)+(LVsR2+L2ω)et/L\displaystyle i(t)= -\frac{LVs}{R^2+ L^2\omega}cos(\omega t)+ \frac{RVs}{R^2+ L^2\omega}sin(\omega t)+ \left(\frac{LVs}{R^2+ L^2\omega}\right) e^{-t/L}.
 
Top