current_voltage_divider_problem1

current_voltage_divider_problem1 - through the circuit and...

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Sample Problem Topic: Voltage & Current Dividers Statement of Problem: Given the circuit shown in the figure below Find a) The value of i a b) The value of i b c) The value of V 0 Solution Of course, we could solve this problem using the previous method(s), namely KVL, KCL. However, we can also re-draw this circuit to quickly determine a few parameters. In certain cases, this can be preferable to solving systems of equations. Redraw the circuit as follows: 50V 16Ω To get the equivalent 16 Ω resistor, simply use the parallel resistor(s) equation: 1 20 + 1 80 ± 1 = 16 Ω We can calculate the current through the series circuit, thus obtaining the amount of current split through the 20 Ω and 80 Ω simply using Ohm’s Law: V = IR V/R = I 50 / (4+16) = 2.5A To find I1, I2, you can now look at the circuit as a current divider. We can do this because we know the total (series) current running
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Unformatted text preview: through the circuit, and we are only concerned with finding the division between R 20 , R 80 . 2.5A 20Ω 80Ω i b i a Simply use the current divider equation to get the final current division. a) ° ? = 2.5 ² ∗ 80 20 + 80 ± = 2 ² b) ° ? = 2.5 ² ∗ 20 20 + 80 ± = .5 ² c) To find the voltage across these parallel resistors, simply remember that parallel resistors will have the same voltage drop. 50V 4Ω 16Ω Returning to the above “equivalent” circuit, we can use a basic voltage divider equation to determ ine the voltage drop across the two resistors: ³ 16 Ω = 50 ³ 16 4 + 16 ± = 40 ³...
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This note was uploaded on 10/30/2011 for the course EE 215 taught by Professor Unknown during the Spring '05 term at University of Washington.

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