MEL TECH
CIRCTRONICS
MESH CURRENT METHOD IN AC CIRCUITS
To reduce the number of equations and unknowns there are two other methods we can use: the node potential and the mesh (loop) current methods. The only difference from DC circuits is that in the AC case, we have to work with complex impedances (or admittances) for the passive elements and complex peak or effective (rms) values for the currents.
In this chapter we will demonstrate these method.
Let's demonstrate the use of the mesh current method.
Example 1
Find the current of the voltage generator V = 10 V, f = 1 kHz, R = 4 kohm, R2 = 2 kohm, C = 250 nF, L = 0.5 H, I = 10 mA, vS(t) = V coswt, iS(t) = i sinw t
Although we could again use the method of node potential with only one unknown, we will demonstrate the solution with the mesh current method.
Let's first calculate the equivalent impedances of R2,L (Z1) and R,C (Z2) to simplify the work:
We have two independent meshes (loops).The first is: vS, Z1 and Z2 and the second: iS and Z2. The direction of the mesh currents are: I1clockwise, I2 counterclockwise.
The two mesh equations are: VS = J1*(Z1 + Z2) + J2*Z2 J2 = Is
You must use complex values for all the impedances, voltages and currents.
The two sources are: VS = 10 V; IS = -j*0.01 A.
We calculate the voltage in volts and the impedance in kohm so we get the current in mA.
Hence:
j1(t) = 10.5 cos (w×t -7.1°) mA
