La correction d'Exercice sur le calcul des tentions et de Résistance et de puissance 6 (Loi des mailles)
1. RT = R1 + R R = R2// R3//( R4 + R5) = 0,5k RT = 1k + 0,5k = 1,5k
2. E = I RT ⇒ I = E / RT = 15V / 1,5k = 10mA
3. U3 = R3I3 = RI = 0,5k x 10mA = 5V
4. U4 = U3 x R4 / (R4 + R5) = 5V x 1,5k / 2k = 3,25V
5. U5 = U3 - U4 = 5V - 3,25V = 1,75V
6. RI = R2I2 = R3I3 = I4(R4 +R5) I2= I3 = 5V / 2k = 2,5mA
2. E = I RT ⇒ I = E / RT = 15V / 1,5k = 10mA
3. U3 = R3I3 = RI = 0,5k x 10mA = 5V
4. U4 = U3 x R4 / (R4 + R5) = 5V x 1,5k / 2k = 3,25V
5. U5 = U3 - U4 = 5V - 3,25V = 1,75V
6. RI = R2I2 = R3I3 = I4(R4 +R5) I2= I3 = 5V / 2k = 2,5mA
et I4 = 5V / 1k = 5mA = I - I2 - I3
7. P1 = R1I²= 1k x (10mA)²= 100mW P2 = P3 = 2k x (2,5mA)²= 12,5mW
8. P4 = 0,75k x (5mA)²= 18,75mW P5 = 0,25k x (5mA)² = 6,25mW
PT =P1 + P2 +P3 + P4 + P5 = 150mW P = EI = 15V x 10mA = 150mW
Conclusion : PT = P
9. Puisque R4 + R5 = R5 ⇒ diminution de la résistance RT ⇒ augmentation de I ⇒ augmentation de PT.
7. P1 = R1I²= 1k x (10mA)²= 100mW P2 = P3 = 2k x (2,5mA)²= 12,5mW
8. P4 = 0,75k x (5mA)²= 18,75mW P5 = 0,25k x (5mA)² = 6,25mW
PT =P1 + P2 +P3 + P4 + P5 = 150mW P = EI = 15V x 10mA = 150mW
Conclusion : PT = P
9. Puisque R4 + R5 = R5 ⇒ diminution de la résistance RT ⇒ augmentation de I ⇒ augmentation de PT.