ELEKTRONIKA ANALOG & DIGITAL

Tugas Persamaan Mesh

Loop 1        = -V + I_R1 . R1 + I_R2 . R2 =0

                       -V + (I₁ – I₃) R₁ + (I₁ – I₂) R₂ = 0

                       -20 + (I₁ – I₃) 100 ­+ ­­­­­(I₁ – I₂) 500 = 0

                       -20 + I₁.100 – I₃.100 + I₁.500 – I₂.500 = 0

                       -20 + I₁.600 – I₃.100 – I₂.500 = 0

                       I₁.600 – I₂.500 – I_3­.100 = 20

Loop 2        = I_R1. R₂ + I_R3 . R₃ + I_R5 . R₅ = 0

                       (I₂ – I₁) R₂ + (I₂ – I₃) R₃ + I₂ . R₅ = 0

                       (I₂ – I₁) 500 + (I₂ – I₃) 100 + I₂ . 200 = 0

                       I₂ . 500 – I₁ . 500 + I₂ . 100 – I₃ . 100 + I₂ . 200 = 0

                       -I₁ . 500 + I₂ . 800 – I₃ . 100 = 0

Loop 3        = I_R4 . R₄ + I_R3 . R₃ + I_R1 . R₁ = 0

                       I₃ .R₄ + (I₃ – I₂) R₃ + (I₃ – I₁) R₁ = 0

                       I₃ . 300 + (I₃ – I₂) 100 + (I₃ – I₁) 100 = 0

                       I₃ . 300  + I₃ . 100 – I₂ . 100 + I₃ . 100 – I₁ . 100 = 0

                       -I₁ . 100 – I₂ . 100 + I₃ . 500 = 0

Jadi,

Loop 1        = I₁.600 – I₂.500 – I_3­.100 = 20

Loop 2        = -I₁ . 500 + I₂ . 800 – I₃ . 100 = 0

LooP3         = -I₁ . 100 – I₂ . 100 + I₃ . 500 = 0

Persamaan dalam bentuk matrixnya adalah :

         600    -500    100         20

         -500  800     -100    =  0

         -100  -100    500         0

        

         a        b         c

         d        e          f     = aei + bfg + cdh + afh + bdi + ceg

         g        h         i

Jadi,

               600     -500    100   

      A =    -500    800     -100

               -100    -100    500

Det(A) = 600*800*500 + (-500*(-100)*(-100)) + ((-100)*(-500)*(-100)) – (600* (-100)*(-100)) – ((-500)*(-500)*500)) – ((-100) *800*(-100)) = 240000000 + (-5000000) + (-5000000) – (6000000) – (125000000) – (8000000) = 240000000 – 5000000 – 5000000 – 6000000 – 125000000 – 8000000 = 9100000

               20       -500    -100

      A1=   0        800     -100

                0        -100    500

Det(A1) = 20*800*500 + (-500*(-100)*0) + (100*0*(-100)) – (20*(-100)*(-100)) – ((-500)*0*500) – (100*800*0) = 8000000 + 0 + 0 – 200000 – 0 – 0 = 7800000

               600     20       100

      A2=   -500    0         -100

               -100    0         500

Det(A2) = 600*0*500 + (20*(-100)*(-100)) + (100*(-500)*0) – (600*(-100)*0) – (20*(-500)*500) – (100*0*(-100)) = 0 + 200000 + 0 – 0 +5000000 – 0 = 5200000

               600     -500    20

      A3=   -500    800     0

               -100    -100    0

Det(A3) =  600*800*0 + ((-500)*0*(-100))  + (20*(-500)*(-100)) – (600*0*(-100)) – ((-500)*(-500)*0) – (20*800*(-100)) = 0 + 0 + 1000000 – 0 – 0 + 1600000 = 2600000

Sehingga,

I1 = Det (A1) / Det (A) = 7800000 / 91000000 = 85,7 mA

I2 = Det (A2) / Det (A) = 5200000 / 91000000 = 57,14 mA

I3 = Det (A3) / Det (A) = 2600000 / 91000000 = 28,57 mA

Maka,

IR1 = I1 –I3 = 85,7 mA – 28,57 mA = 57,13 mA

IR2 = I1 – I2 = 85,7 mA – 57,14 mA = 28,56 mA

IR3 = I2 – I3 = 57,14 mA – 28,57 mA = 28,57 mA

IR4 = I3 = 28,57 mA

IR5 = I2 = 57,14 mA

VR1 = IR1*R1 = 57,13 mA*100 Ω = 5,713 v

VR2 = IR2*R2 = 28,56 mA*500 Ω = 14,280 v

VR3 = IR3*R3 = 28.57 mA*100 Ω = 2,857 v

VR4 = IR4*R4 = 28,57 mA*300 Ω = 8,571 v

VR5 = IR5*R5 = 57,14 mA*200 Ω = 11,42 v