Mexico

205

INSTITUTO POLITECNICO NACIONAL

Centro de Estudios Científicos y Tecnológicos No. 4

Santiago Gómez Juan José

FÍSICA III

EXAMEN DELEYES DE KIRCHHOFF

5IV3

R1= 7 Ω
R2= 9 Ω
R3= 15 Ω
ε1= 100 V
ε2= 70 V ΣE = ΣS
ε3= 95 V 0 = I1 + I2 + I3
ε4= 55 V I1 +I2 + I3 = 0 A
r1= 0.3 Ω
r2= 0.6 Ω
r3= 0.9 Ω
r4= 0.5 Ω


MALLA 1

Σε = ΣIR
ε1 – ε2 + ε3 = – I1R1 – I1r1 + I2r2 + I2R2 +I2r3
ε1 – ε2 + ε3 = – I1 [R1 + r1] + I2 [r2 + R2 + r3]
100 – 70 + 95 = – I1 [7 + 0.3] +I2 [0.6 + 9 + 0.9]
- 7.3 I1 + 10.5 I2 = 125 B

MALLA 2

Σε =ΣIR
– ε2 + ε3 – ε4 = I2r2 + I2R2 + I2r3 – I3R3 – I3r4
– ε2 + ε3 – ε4 = I2 [r2 + R2 + r3] – I3 [R3 + r4]
- 70 + 95 – 55 = I2 [0.6 + 9 + 0.9] – I3 [15 +0.5]
10.5 I2 – 15.5 I3 = - 30 C

I1 + I2 + I3 = 0 A
- 7.3 I1 + 10.5 I2 = 125 B
10.5 I2 – 15.5 I3 = - 30 C
Despejar I3 de Ec. A
I3 = - I1 - I2Sustituir I3 de Ec. A en Ec. C
10.5 I2 – 15.5 [- I1 – I2] = - 30
10.5 I2 + 15.5 I1 + 15.5 I2 = - 30
26 I2 + 15.5 I1 = - 30 D
Despejar I1 de Ec. B yD
- 7.3 I1 + 10.5 I2 = 125
10.5 I2 = 125 + 7.3 I1
I1 = (10.5 I2 - 125) / 7.3 E
I1 = (- 26 I2 – 30) / 15.5
Igualar I1
(10.5 I2 - 125) / 7.3 = (-26 I2 - 30) / 15.5
15.5 (10.5 I2 - 125) = 7.3 (- 26 I2 - 30)
162.75 I2 – 1937.5 = - 189.8 I2 – 219
162.75 I2 + 189.8 I2 = - 219 + 1937.5
352.55 I2 =1718.5
I2 = 1718.5 / 352.55 = 4.784 A
Sustituir I2 en E
I1 = (10.5 (4.874) - 125) / 7.3
I1 = - 10.1127 A
Sustituir I1 e I2 en A
I3 = 5.238 A