Unformatted text preview: Redraw the circuit as follows: 500Ω .3A 2kΩ i 2 i 1 We can calculate the current through the parallel circuit, thus obtaining the amount of current split through the 2k Ω and 500Ω resistors Simply use the current divider equation to get the final current division. ² 1 = .3 ³ ∗ 500 2000 + 500 ± = − .06 ³ ´µ − 60 ¶³ ² 2 = .3 ³ ∗ 2000 2000 + 500 ± = − .24 ³ ´µ − 240 ¶³ Since the current source is pointing downwards, this defines the convention of the current flow. When looking at the top node, you will notice that the arrows of i1 and i2 are pointing “outwards.” Notice that the positive current direction is outwards from the top node as well. Since all currents going in or out of a node sum to zero, it explains why the currents are negative....
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 Spring '05
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 Electrical Engineering, Volt, Voltage divider, Resistor, Electrical impedance, dependent current source

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