Use PSpice to find i1, i2, and i3 in the circuit displayed in Fig. 13.53.

Question

Use PSpice to find [latex]i_1, i_2,[/latex] and [latex]i_3[/latex] in the circuit displayed in Fig. 13.53.

Accepted Answer

The coupling coefficients of the three coupled inductors are determined as follows:

[latex] k_{12}=\frac{M_{12}}{\sqrt{L_1 L_2}}=\frac{1}{\sqrt{3 imes 3}}=0.3333 \ \space \ k_{13}=\frac{M_{13}}{\sqrt{L_1 L_3}}=\frac{1.5}{\sqrt{3 imes 4}}=0.433 \ \space \ k_{23}=\frac{M_{23}}{\sqrt{L_2 L_3}}=\frac{2}{\sqrt{3 imes 4}}=0.5774 [/latex]

The operating frequency f is obtained from Fig. 13.53 as ω = 12 π = 2 πƒ → ƒ = 6 Hz.
The schematic of the circuit is portrayed in Fig. 13.54. Notice how the dot convention is adhered to. For L2, the dot (not shown) is on pin 1 (the left-hand terminal) and is therefore placed without rotation. For L1, in order for the dot to be on the right-hand side of the inductor, the inductor must be rotated through 180°. For L3, the inductor must be rotated through 90° so that the dot will be at the bottom. Note that the 2-H inductor ([latex]L_4[/latex]) is not coupled. To handle the three coupled inductors, we use three K_LINEAR parts provided in the analog library and set the following attributes (by double-clicking on the symbol K in the box):

K1 - K_LINEAR
L1 = L1
L2 = L2
COUPLING = 0.3333

K2 - K_LINEAR
L1 = L1
L2 = L3
COUPLING = 0.433

K3 - K_LINEAR
L1 = L2
L2 = L3
COUPLING = 0.5774

[latex]\fbox{K}[/latex] K1
K_Linear
COUPLING = 0.3333
L1 = L1
L2 = L2

[latex]\fbox{K}[/latex] K2
K_Linear
COUPLING = 0.433
L1 = L2
L2 = L3

[latex]\fbox{K}[/latex] K3
K_Linear
COUPLING = 0.5774
L1 = L1
L2 = L3

Three IPRINT pseudocomponents are inserted in the appropriate branches to obtain the required currents [latex]i_1, i_2, ext{ and } i_3[/latex]. As an AC single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts =  1, Start Freq =  6, and Final Freq =  6. After saving the schematic, we select Analysis/Simulate to simulate it. The output file includes:

FREQ IM(V_PRINT2) IP(V_PRINT2)
6.000E+00 2.114E-01 -7.575E+01
FREQ IM(V_PRINT1) IP(V_PRINT1)
6.000E+00 4.654E-01 -7.025E+01
FREQ IM(V_PRINT3) IP(V_PRINT3)
6.000E+00 1.095E-01 1.715E+01

From this we obtain

[latex] \pmb{I _1}=0.4654 \underline{/-70.25^{\circ}} \ \space \ \pmb{I _2}=0.2114 \underline{/-75.75^{\circ}}, \quad \pmb{I _3}=0.1095 \underline{/ 17.15^{\circ}}[/latex]

Thus,

[latex]i_1 = 0.4654 \cos (12 πt − 70.25^{\circ}) A\ \space \ i_2 = 0.2114 \cos(12 πt − 75.75^{\circ}) A \ \space \ i_3 = 0.1095 \cos(12 πt + 17.15^{\circ}) A[/latex]

Question 13.13: Use PSpice to find i1, i2, and i3 in the circuit displayed i...

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