PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB
250 Practical MATLAB® Applications for Engineers The calculations for the currents I 1 , I 2 , and I 3 are verifi ed using MATLA ...
Alternating Current Analysis 251 and X jC j j CC Yj 11 10 3 5 1 6 6 ()( ) The circuit of Figure 3.23 is then redrawn w ...
252 Practical MATLAB® Applications for Engineers Solving for E 1 and E 2 by hand yields E 1 50 5 03 10 5 5 53 15 ...
Alternating Current Analysis 253 R.3.61 The example shown in Figure 3.25 uses the Thevenin’s theorem to calculate the cur- rent ...
254 Practical MATLAB® Applications for Engineers and Zj jj jj TH j 14 8484 8484 64 ()() (by replacing the voltag ...
Alternating Current Analysis 255 ANALYTICAL Solution Replacing ZL by a short (Norton’s theorem) results in the circuit diagram s ...
256 Practical MATLAB® Applications for Engineers Then, ZTH = [(Z 1 ||Z 2 ) + Z 3 ] ||Z 4 Ω Z||Z Z jj 12 3 jj ()() .. ...
Alternating Current Analysis 257 R.3.64 The superposition principle for the AC case follows closely to the DC case discussed in ...
258 Practical MATLAB® Applications for Engineers Now consider t he AC source by set t i ng t he DC source to zero (VA = 0 ), and ...
Alternating Current Analysis 259 Then the currents of the circuit diagram shown in Figure 3.34 are evaluated by the algebraic ad ...
260 Practical MATLAB® Applications for Engineers R.3.67 Let the load impedance of an arbitrary network be ZL = RL + jXL, where R ...
Alternating Current Analysis 261 R.3.69 The transformation from ∆ to Y is accomplished by the following set of equations: Z ZZ A ...
262 Practical MATLAB® Applications for Engineers R.3.77 3 Φ Systems are the most common confi gurations used to generate and del ...
Alternating Current Analysis 263 R.3.80 The system voltages, as well as the loads in a 3 Φ system can be connected in either a ∆ ...
264 Practical MATLAB® Applications for Engineers R.3.84 The magnitude of the line currents in a balanced 3 Φ system, as well as ...
Alternating Current Analysis 265 R.3.86 The example shown in Figure 3.47 illustrates the general approach employed to solve for ...
266 Practical MATLAB® Applications for Engineers Then, I 1 = I 12 − I 31 I 1 12 ∠∠ ∠ 90 12 3012 3 120 A and I 2 = I 23 ...
Alternating Current Analysis 267 Then, I 12 120 0 30 90 490 ∠ ∠ ∠ I 23 120 120 30 90 4 150 ∠ ∠ ∠ I 31 120 12 ...
268 Practical MATLAB® Applications for Engineers C = 1e-6; w = [200:50:2000]; XL = w*L; XC = 1./(w*C); plot(w,XL,’o’,w, XC,’*’) ...
Alternating Current Analysis 269 ANALYTICAL Solution XC() = −j 5 XL() = j 2 XXX XX XX jj CL jj j CL CL () ()//() () () () () ...
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