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Lecture - Current Divider Circuit Current source and two...

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1/21/14 1 Current Divider Circuit I s i 1 i 2 R 1 R 2 V Current source and two resistors connected in parallel KCL: I s = i 1 + i 2 V = I s R 1 R 2 R 1 + R 2 I s = V R 1 + V R 2 I s = V R 2 + R 1 R 1 R 2 i 1 = ? i 2 = ? i 1 = V R 1 i 1 = I s R 2 R 1 + R 2 i 2 = V R 2 i 2 = I s R 1 R 1 + R 2 i 1 i 2 = R 2 R 1 Example (2) 10V 2 k Ω 10 k Ω R x v o In the circuit shown, select a value for the resistor R x so that v o =8V
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1/21/14 2 Solution KVL (abca): 10 = v + 8 v = 2 V i = 2 2 k = 1 mA Ohm s Law: i o = 8 10 k = 0.8 mA KCL node b: i = i x + i o 10V 2 k Ω 10 k Ω R x v o i v a b c c c i o i x 1 m = i x + 0.8 m i x = 0.2 mA Current division rule: i o i x = R x 10 k 0.8 0.2 = R x 10 k R x = 40 k Ω Given: v o = 8 V Example (3) V S R 2 R 1 R 3 v o Use the voltage division rule to find the output voltage v o
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1/21/14 3 Solution: V S R 2 R 1 R 3 v o current through R 3 is zero ! a b c b v o Hint: The voltage source and R 1 and R 2 connected in series ! Voltage division rule: v o = V s R 2 R 1 + R 2 5A 20 Ω 20 Ω 5 Ω 5 Ω i x i y i z Example (4) Find the current i x Find the current i y Find the current i z
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