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Solve Current Through 3 Ohm Resistor: Kirchoff's Rule & I=V/R
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[QUOTE="deamonata, post: 4334949, member: 470526"] So the best way to answer this without just giving you the answer it to solve a similar question so with that in mind, I will use this circuit. [ATTACH=full]160373[/ATTACH] There are other methods out there but I'm going to use the Branch-Current method (look up loop current method). Step 1 is to define the branch currents and the direction that they are acting in. Steb 2 Define the voltages in terms of the currents [tex]V_{ab} = 2I_1[/tex] [tex]V_{bd} = 8I_3[/tex] [tex]V_{cb} = 4I_2[/tex] Step 3 using Kirchoffs Voltage law we get [tex]ƩV_{abda} = 0 = V_{ab} + V_{bd} + V_{da}[/tex] [tex]ƩV_{bcdb} = 0 = V_{cb} + V_{bd} + V_{dc}[/tex] These can be re written as [tex]ƩV_{abda} = V_{ab} + V_{bd} + V_{da} = 2I_1 + 8I_3 - 32 [/tex] [tex]ƩV_{bcdb} = V_{cb} + V_{bd} + V_{dc} = 4I_2 + 8I_3 - 20 [/tex] then finally You use Kichchoffs current laws at node b to get your final equation. [tex]ƩI_b = 0 = I_1 + I_2 - I_3[/tex] this gives the 3 simultaneous equations [tex]I_1 + I_2 - I_3= 0[/tex] [tex]2I_1 + 8I_3 = 32 [/tex] [tex]4I_2 + 8I_3 = 20 [/tex] Solving gives the currents [itex]I_1 = 4[/itex], [itex]I_2 = -1[/itex] and [itex]I_3= 3[/itex]. Hope that helps you. [/QUOTE]
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Solve Current Through 3 Ohm Resistor: Kirchoff's Rule & I=V/R
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