R₁ 5.0 Ω b R3 78 Ω E₁ = 24.0 V 11 T4₁ a 12 с 0.10 Ω E₂ = 48.0 V 0.50 Ω E3 = 6.0 V 13 i 0.05 92 h R₂ mu 40 Ω R₁ • 20 Ω e *0.20 Ω g E4 = 36.0 V a: Apply the junction rule to junction "a" shown in the Figure and write down the resulting equation. b: Apply the loop rule to loop abcdefgha in the Figure and write down the resulting equation. c: Apply the loop rule to loop aedcba in the Figure and write down the resulting equation. d: Solve for the currents 1₁, 12 and 13.

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Chapter1: Units, Trigonometry. And Vectors
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Use Kirchoff's Rules to solve for all of the currents in the circuit shown below.

R₁
5.0 22
a
R3
7892
£₁ = 24.0 V
Hlad
с
0.10 Ω
b
12
13
E₂ = 48.0 V
12
0.50 Ω
E3 = 6.0 V
13
From
R₂
40 92
R5
- 20 Ω
e
14
*0.20 Ω
g
E4 = 36.0 V
a: Apply the junction rule to junction "a" shown in the Figure and write
down the resulting equation.
b: Apply the loop rule to loop abcdefgha in the Figure and write down
the resulting equation.
c: Apply the loop rule to loop aedcba in the Figure and write down the
resulting equation.
d: Solve for the currents 1₁, 12 and 13.
Transcribed Image Text:R₁ 5.0 22 a R3 7892 £₁ = 24.0 V Hlad с 0.10 Ω b 12 13 E₂ = 48.0 V 12 0.50 Ω E3 = 6.0 V 13 From R₂ 40 92 R5 - 20 Ω e 14 *0.20 Ω g E4 = 36.0 V a: Apply the junction rule to junction "a" shown in the Figure and write down the resulting equation. b: Apply the loop rule to loop abcdefgha in the Figure and write down the resulting equation. c: Apply the loop rule to loop aedcba in the Figure and write down the resulting equation. d: Solve for the currents 1₁, 12 and 13.
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