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Calculate the three currents I1, I2, and I3 indicated in the circuit diagram shown in Fig. P26.59.

Short Answer

Expert verified

The three currents areI1=0.848A,I2=2.14A,andI3=0.171A

Step by step solution

01

Concept Introduction

02

Given data

E1=12.0VE2=9.0VR1=1.00ΩR2=5.00Ω

03

Apply Kirchoff’s loop law

Use loop rule to loop (1),

-E1+I2R1+(I2-I3)R2=0

-E1+I2(R1+R2)-I3R2=0 (1)

For loop 2,

E2-(I1+I3)R3-I1R1=0

E2-I3R3-I1(R1+R3)=0 (2)

Loop 3,

-E1+I2R1-I1R1+E2+I3R4=0 (3)

04

Use the values in the above expressions

Put R2= 5 R1, R3= 8 R1, R4= 10 in equations 1,2, and 3, and we get,

localid="1668335255112" -E1+6I2R1-5I3R1=0E2-8I3R1-9I1R1=0-E1+I2R1-I1R1+E2+10I3R1=0

Divide the above equation by R1,

E1R1+6I25I3=06I25I3=E1R14

localid="1668330342001" E2R18I39I1=08I3+9I1=E2R15

E1R1+I2I1+E2R1+10I3=0I2I1+10I3=E1R1E2R16

Use the values from equations (4), and (6) and we get,

5I36+E16R1I1+10I3=E1R1E2R1

Multiply by factor 6, and we get,

5l36l1+60I3=5E1R16E2R16l1+65l3=5E1R16E2R1 (7)

Multiply the equation (5) by factor 2/3,

16l33+6l1=2E23R1

Add the above expression from equation (7), and we get

16I33+65I3=2E23R1+5E1R16E2R1211I33=15E116E23R1I3=15E116E2211R1

05

Calculate the value of currents 

Substituting all the values we get,

I3=15E116E2211R1=15(12.0V)16(9.00V)211(1.00Ω)=0.17A

from this equation substitute the second one in (5), and we get,

I1=E29R18I39=9.00V9(1.00Ω)8(0.171A)9=0.848A

Substitute to the first equation in (4),

I2=5(0.171A)6(12.0V)6(1.00Ω)=2.14A

Thus, I1=0.848A,I2=2.14A,andI3=0.171A.

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