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Find the currents flowing in the circuit in the figure below. (Assume the resistances are
R1 = 8 Ω, R2 = 10 Ω, R3 = 12 Ω, R4 = 15 Ω, r1 = 0.5 Ω, r2 = 0.25 Ω, r3 = 0.75 Ω, and, r4 = 0.25 Ω.)

Find the currents flowing in the circuit in the figure below Assume the resistances are R1 8 Ω R2 10 Ω R3 12 Ω R4 15 Ω r1 05 Ω r2 025 Ω r3 075 Ω and r4 025 Ω class=

Respuesta :

By Kirchoff's law in left side loop

[tex]E_1 + E_2 = {r_1 + R_1 + R_4)I_1 + (R_2 + r_2) I_3[/tex]

similarly kirchoff's law in right side loop

[tex]E_3 - E_4 - E_2 = (r_3 + r_4 + R_3)I_2 - (R_2 + r_2)I_3[/tex]

also by junction law we know that

[tex]I_1 = I_2 + I_3[/tex]

now by plug in all values we have

[tex]18 + 3 = (0.5 + 8 + 15)I_1 + (10 + 0.25)I_3[/tex]

[tex]21 = 23.5I_1 + 10.25I_3[/tex]

[tex]12 - 24 - 3 = (0.75 + 0.25 + 12)I_2 - (10+ 0.25)I_3[/tex]

[tex]-15 = 13I_2 - 10.25I_3[/tex]

So by solving above equations we have

[tex]I_1 = 0.492 A[/tex]

[tex]I_2 = -0.428 A[/tex]

[tex]I_3 = 0.920 A[/tex]

batolisis

The currents flowing in the circuit are as follows :

  • I₁ = 0.492 A
  • I₂ = -0.428 A
  • I₃ = 0.920 A

Determine the currents in the circuit

Given the various resistances and circuit we will Kirchoff's law and Junction Law

First step : Apply Kirchoff's law to the left side loop

E₁ + E₂ = ( r₁ + R₁ + R₄ ) I₁ + ( R₂ + r₂ ) I₃

18 + 3 = ( 0.5 + 8 + 15 ) I₁ + ( 10 + 0.25 ) I₃

21 = 23.5 I₁ + 10.25 I₃ --- ( 1 )

Next step : apply Kirchoff's law to the right side loop

E₃ - E₄ - E₂ = ( r₃ + r₄ + R₃ ) I₂ -  ( r₂ + R₂ ) I₃

12 - 24 - 3 = ( 0.75 + 0.25 + 12 ) I₂  -  ( 0.25  + 10 ) I₃

-15 = 13 I₂ - 10.25 I₃ --- ( 2 )

Note : Applying the law of Junction : I₁ = I₂ + I₃

Resolve equations ( 1 ) and ( 2 ) simultaneously

I₁ = 0.492 A,  I₂ = -0.428 A,  I₃ = 0.920 A

Hence we can conclude that The currents flowing in the circuit are as follows : I₁ = 0.492 A,  I₂ = -0.428 A,  I₃ = 0.920 A.

Learn more about Kirchoff's law : https://brainly.com/question/86531

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