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Please answer the following question and show all work:

Problem 1: For the circuit shown below, use Kirchhoffs laws to calculate the currents 1, 12, 13. 40 1 40 20 V 2V 10V w OE OZ
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Answer #1

applying kirchoff's voltage law in left loop

so -20 + 4I_{1}+ 2+3I_{2}+4I_{1}=0

8I_{1}+3I_{2}=18 -----(1)

applying kirchoff's voltage law in right  loop

-10 + 4I_{3}+2+3I_{2}+2I_{3}= 0

613 +312 = 8 - - - - (2

applying kirchoff's current   law at upper node

so I_{1}+I_{3}= I_{2}--------------(3)

now put the value of  I_{2} from  equation (3) in equation (1) & equation (2)

so the modified eqns

3I_{1}+9I_{3}= 8-------(4)

11I_{1}+3I_{3}=1 8-------(5)

by solving eqn(4) &eqn(5) we got

I_{1}= 1.5333A

I_{3}= 0.378A

put the value of I_{3} and I_{1} in eqn (3)

so we got I_{2}= 1.9111A

so

I_{1}= 1.533A

I_{2}= 1.9111A

I_{3}= 0.377A

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