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Find the current through 12 Omega resist...

Find the current through `12 Omega` resistor in

Text Solution

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Let V be the potential at P, then applying KCL at junction
P, we get
`I = I_1+I_2+I_3 rArr (15-V)/(12) = (V -2)/(6) + (V-3)/(4) + (V-4)/(8)`
or `V = (68)/(15)V`
and `I = (15 - (68 //15))/(12) = (157)/(180)A`
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