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

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Let current in the circuit be distributed like,

where $I_1$ , $I_2$ and $I_3$ are the different currents through shown branches.

Now, Applying KVL in the Loop,

$10-I10-I_25-(I_2+I_3)10=0$

Also, we have $I=I_1+I_2$

so putting it in the KVL equation, we get

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

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

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

Now let's apply KVL in the loop involving $I_1$ , $I_2$ and $I_3$ ,

$5I_2-10I_1-5I_3=0$ .................................(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 equations and 3 variables, on solving we get,

$I_1= \frac{4}{17}A$

$I_2= \frac{6}{17}A$

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

And, the total Current in the circuit is,

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

Posted by

Pankaj Sanodiya

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