3.9 Determine the current in each branch of the network shown in Fig. 3.30:

            

Answers (1)

Let current in the circuit is distributed like 

where I1, I2, and I3 are the different current through shown branches.

Now, Applying KVL in Loop

 10-I10-I_25-(I_2+I_3)10=0

Also, we have I=I_1+I_2

so putting it in kvl equation

10-(I_1+I_2)10-I_25-(I_2+I_3)10=0

10-10I_1-10I_2-5I_2-10I_2-10I_3=0

10-10I_1-25I_2-10I_3=0.................................(1)

Now let's apply kvl in the loop involving I1 I2 AND I3

5I_2-10I_1-5I_3=0.................................(2)

now, the third equation of KVL 

5I_3-5(I_1-I_3)+10(I_2+I_3)=0

-5I_1+10I_2+20I_3=0..............................(3)

Now we have 3 equation and 3 variable, on solving we get

I_1= \frac{4}{17}A

I_2= \frac{6}{17}A

I_3= \frac{-2}{17}A

Now the total Current 

I=I_1+I_2=\frac{4}{17}+\frac{6}{17}=\frac{10}{17}

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