# Mesh Current Method Circuit Analysis For AC Transient and Steady State Solution of RL Circuit Hi everyone, I'm trying to find the currents through each branch for the above circuit using mesh current method (as I don't really know how else to do it), and end up with two simultaneous differential equations:

$$Mesh 1: R_si_1+R_1(i_1-i_2)+L_1\frac{d}{dt}(i_1-i_2)=155cos(377t)$$ $$Mesh 2: R_2i_2+L_2\frac{d}{dt}i_2+L_1\frac{d}{dt}(i_2-i_1)+R_1(i_2-i_1)=0$$

These simplify to:

$$Mesh 1: (R_s+R_1)i_1-R_1i_2+L_1\frac{d}{dt}i_1-L_1\frac{d}{dt}i_2=155cos(377t)$$ $$Mesh 2: -R_1i_1+(R_1+R_2)i_2-L_1\frac{d}{dt}i_1+(L_1+L_2)\frac{d}{dt}i_2=0$$

The problem I am having is that I cannot figure out a linear combination of these equations to eliminate either i_1 or i_2 variables to solve the system.

Am I going about this the wrong way? Or is there an easier way to solve this?

I can readily find the steady state solution reducing everything into impedances but I need to find the transient solution as well.

## 1 Answer

It would be easier to convert both the differential equations to laplace domain and then solve for i1(s) and i2(s). Take inverse laplace then to find i1(t) and i2(t).