Question

QUESTION 7 Using the circuit below, answer the following questions: -20 1Ω M 312 22 -14.2 we Ri Zci 122 ZLI R3 Żc2 202 ZL2 V 32/ºv R2320 1 1. What is the current / 1 in polar format? Answer Format: (1=XZYºA Y = Answer: X = Example: If your calculated current is 0.213 2143 then X = 2.13e-1, Y = 1.43e2 Note: Adjust the phase angle such that - 180° <Y< 180° 2. What is the current loin polar format? Answer Format: = XZYº A Answer: X = Y = Example: If your calculated current is 0.213 2143° then X = 2.13e-1, Y = 1.43e2 Note: Adjust the phase angle such that - 180° <Y< 180°

Answer 1

The circuit is:

20-14 10 -22 302 w R 32 v RZCI 202 20 12 I1 ZLI ZL2 Mesh-1 Mesh-2 R32 11

Consider the circuit as shown in the figure, in the mesh-1, the current enters the dot end while in the mesh-2, the current comes out of the dot end. Hence mutual inductance M will be negative or M (-ve).

Apply the mesh analysis in the Mesh-1:

(2-j4)/2 + 121 + 2(11-1) + (-31)

3220° = 211 - 0414 +221 +211 - 21. - jl.

3220º = 1/(4 - j2) - 1.(2+j)... ........ (1) I..

Apply the mesh analysis in the Mesh-2:

0 = (3- j2)1, + 2(10-11) + 21, + (-21)

0 = 316 - 221, +21. – 21 +j21- jlly

0 = 11(-2-j1) + 510 ........(2)

(2+j1)11 1. (3) 5

Put the equation (3) into (1)

3220º = 1/(4 - 12) - (2+011 (2+j) 5

32 = (20 - 010 - (4 + 12 + j2 - 1) 5 j2 – 1)) 1

32 = (20 – 310 – 4-24 + 1) 5 j4 + ) 1.2

11 160 17 - 314

160 11 V172 + 142 Z360° – tan-1 14 17

11 160 22.02_320.539

11 = 7.2662 320.53°

11 = 7.2662360° – 320.530

In=7.266239.470

I1 convert into the required form:

11 = XZYº

X = 0.7266e1

Y = 3.947e1

Put the I1 into equation (3)

10 (2 + j1) x 7.266239.47° 5

V22 + 12 _tan-- () x 7.266_39.47° 1. 5

10 2.24226.57° x 7.266239.47° 5

16.276266.04° 10 5

10 = 3.254266.04°

I0 convert into the required form:

To = XZY*

X = 0.3254e1

Y = 6.604e1