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Hi anyone. I try to de-draw the circuit in the simpler to calculate i1 and i2. But can not. Anyone have any hint to do this question ? . Or just give me the specific name of the way of how to do that. :( thanks

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3 Answers 3

Use kirchoff's loop and junction law. :)

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The total resistance seen by the voltage source is: $$ R_{tot}=[(6//12)+(12//6)]\Omega = 8\Omega $$ then: $$ i_1=\frac{20V}{8\Omega}=2.5A $$

Now let's calculate the voltage of the intermediate node, i.e. the node where you measure $i_2$: $$ V_{mid}=20V\frac{12//6}{(12//6)+(12//6)}=20V\times\frac12=10V $$ The current $i_2$ can be computed as the current in the top $6\Omega$ resistor minus the current in the bottom $12\Omega$ resistor, i.e.:

$$ i_2 = \frac{10V}{6\Omega} - \frac{10V}{12\Omega} = 833mA $$

and there you go.

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try re-drawing it like this:

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then work out the total resistance in the circuit and use V-IR to calculate i1. then find the percentage of current goung through each branch of the wire.

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protected by Qmechanic Apr 20 at 13:24

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