PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

236 Practical MATLAB® Applications for Engineers

ANALYTICAL Solution

XL = jL = j (^1000) 0.02 = j20

X
j
C
j
C
j






 1000 20 10
6 50

• 
  
  
  
  • 
    
    X = XL + XC = j20 − j50 = −j30
    
    Then, Z = R + X = (20 − j30)
    and
    G
    Zj
    j
    
    
    j
    
    
    
    
    
    
    
     
    11
    20 30
    20
    20 30
    30
    20 30
    2
    130
    3
    22 22 130
     1
    
    
    
    (or siie)
    R.3.35 For example, evaluate the voltage v(t) across the series RL connection shown in
    Figure 3.3, assuming that the current through is given by i(t) = Im sin(t).
    ANALYTICAL Solution
    v(t) = (^) √

    ---

    
    

    R^2 + (L)^2 I (^) m sin(t + )
    where θ = tan−^1 (ωL/R).
    The phase angle θ is positive since the voltage leads the current in an RL circuit
    (similar to R.3.31).
    The phasor diagram is shown in Figure 3.4.
    R.3.36 Let us explore now the parallel RL case. Compute the current i(t) for the parallel RL
    circuit diagram shown in Figure 3.5, assuming that the applied voltage is given by
    v(t) = Vm cos(ωt)
    ANALYTICAL Solution
    The total admittance Y is given by
    YG
    L
    
    ^2 
    2
    (^11)
    
    
    
    
    
    
    FIGURE 3.3
    Circuit diagram of R.3.35.

    
    
  
  
  
  

• R –


• L –
  vR(t) i(t)
  vL(t)
  v(t)


• –
  FIGURE 3.4
  Phasor diagram of R.3.35.
  = R
  ωL
  θ tan−^1
  VL
  IL