Find the value of currents $l_{1}, l_{2} a n d l_{3}$

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Solution

since all arms are connected between two nodes, hence PD across both nodes will be same and current through 6 $Ω$ resistance will be $I = I_{1} + I_{2} + I_{3}$
Now, PD across nodes through each branch will be
$V = 6 I = 8 - I_{1} = 9 - 2 I_{2} = 10 - 3 I_{3}$
thish gives
$I_{1} = 8 - V$ ...(1)
$I_{2} = \frac{9 - V}{2}$ ...(2)
$I_{3} = \frac{10 - V}{3}$ ...(3)
solving for $I = I_{1} + I_{2} + I_{3}$

we get, $I = (8 + \frac{9}{2} + \frac{10}{3}) - (1 + \frac{1}{2} + \frac{1}{3}) V = \frac{95}{6} - \frac{11}{6} V$

and using it in $V = 6 I = 95 - 11 V$ , we get value of $V = \frac{95}{12} = 6$ volt
using value of V in eq. 1 , 2 and 3 we get value of $I_{1} = 2 A, I_{2} = 1.5 A, I_{3} = 1.33$ A

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