Transformers: Solving for i_1 and i_2 with Primary and Secondary Coils

[John 123]

Homework Statement

Primary coil of a transformer with an emf E(t),resistance R_1 and inductance L_1.Secondary coil resistance R_2 and inductance L_2.
Let i_1 and i_2 be the respective currents in each coil.

Homework Equations

[tex]
L_1\frac{di_1}{dt}+R_1i_1+M\frac{di_2}{dt}=E(t)
[/tex]
and
[tex]
L_2\frac{di_2}{dt}+R_2i_2+M\frac{di_1}{dt}=0
[/tex]
where M =mutual inductance.
Solve the above system if E(t)=0
[tex]
M^2<L_1L_2
[/tex]

The Attempt at a Solution

I have solved for i_1[which agrees with the book answer].
[tex]
i_1=C_1e^{at}+C_2e^{bt}
[/tex]
where
[tex]
a,b=\frac{-(L_2R_1+L_1R_2)+/_\sqrt{(L_2R_1-L_1R_2)^2+4M^2R_1R_2}}{2(L_1L_2-M^2)}
[/tex]
However I cannot get the correct answer for i_2, which is as follows:
[tex]
i_2=\frac{1}{MR_2}[(L_1L_2-M^2)(aC_1e^{at}+bC_2e^{bt})+L_2R_1(C_1e^{at}+C_2e^{bt}]
[/tex]
I used the method of solving the characteristic equations in m.
Thus for i_1 and i_2 both have the same characteristic equation:
[tex]
(L_1L_2-M^2)m^2+(L_1R_2+L_2R_1)m+R_1R_2=0
[/tex]
Thus we may write
[tex]
i_1=C_1e^{at}+C_2e^{bt}
[/tex]
and
[tex]
i_2=k_1e^{at}+k_2e^{bt}
[/tex]
where a,b are as defined above.
The problem arises when I write k_1 and k_2 in terms of C_1 and C_2 which can be achieved by substituting i_1 and i_2 solutions back into the first equation.
And although I can manipulate the result to be similar to the book answer for i_2 I cannot get complete agreement.
Help!
John

 

[Jhero]

Dear John,

Thank you for your question. It seems like you have made some progress in solving the system of equations for the currents in the primary and secondary coils of the transformer. However, it appears that you are having some trouble with the final step of finding the solution for i_2.

Firstly, it is important to note that in order to solve for i_2, you will need to use the values of a and b that you have already calculated for i_1. This is because the equations for i_1 and i_2 are coupled together through the mutual inductance term, M, and therefore, the values of a and b will be the same for both equations.

With that in mind, let's take a closer look at the equation you have for i_2:

i_2=\frac{1}{MR_2}[(L_1L_2-M^2)(aC_1e^{at}+bC_2e^{bt})+L_2R_1(C_1e^{at}+C_2e^{bt}]

In order to simplify this expression, we can use the fact that a and b are complex conjugates of each other (due to the +/- sign in the numerator of the expression for a and b). This means that we can rewrite the terms in the parentheses as follows:

aC_1e^{at}+bC_2e^{bt}=e^{at}(C_1a+C_2b)

Now, if we substitute this into the equation for i_2, we get:

i_2=\frac{1}{MR_2}[(L_1L_2-M^2)e^{at}(C_1a+C_2b)+L_2R_1(C_1e^{at}+C_2e^{bt}]

Next, we can use the fact that a and b are complex conjugates again to rewrite the term (C_1a+C_2b) as follows:

C_1a+C_2b=(C_1+C_2)a

Now, if we substitute this into the equation for i_2, we get:

i_2=\frac{1}{MR_2}[(L_1L_2-M^2)e^{at}(C_1+C_2)a+L_2R_1(C_1e^{at}+C