Solutions d'exercices des Circuits en sinusoidal- Circuits électriques

Solution d’exercice 1

Donc

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Solution d’exercice 2

1. i = 15,5A sin (100π t - π/6) = Imax sin (100π t - π/6)                 Imax = 15,5A

2. I_eff = Imax / √2 = 15,5A/√2                 Ieff = 10,96A

3. i = 15,5A sin (100π t - π/6) = 15,5A sin (ωt - π/6) ω =100π ω = 628rad/s ω =100π = 2πf 100 = 2f                f = 50Hz T = 1/f          T = 20ms

4. i = 15,5A sin (100π t - π/6) = 15,5A sin (2πf t - π/6) = 15,5A sin (2π t/T - π/6) t = 0 i = 15,5A sin (0 - π/6)

i = -7,75A t = 5ms = T/4 i = 15,5A sin (2π T/4T - π/6) = 15,5A sin (π/3)

i = 13,42A t = 10ms = T/2 i = 15,5A sin (2π T/2T - π/6) = 15,5A sin (5π/6)

i = 7,75A

5. U_R = R_i= 20Ω x 15,5A sin (100π t - π/6)      U_R = 310V sin (100π t - π/6)

6. U_Reff = 310V/ √2              U_Reff =219,2V 

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