Tutorial 2 - parallel R R Therefore, R eq = 14 + 4 = 18 .

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F i n a l M a r k : ECTE 290 Tut 2 Student Name: All questions 1 mark unless indicated. Total 10 Marks. 1. Open circuit is not allowed for a current source. 2. When current enters a resistor from the negative end of reference voltage polarity the Ohm’s law has the form v=-iR 3. When two resistors are in parallel connection they have the same voltage. 4. (4 marks) Find the equivalent resistance of the circuit. Solution: Known quantities: Schematic of the circuit. Find: The equivalent resistance of the circuit. Analysis: Starting from the right side, we combine the two parallel resistors, namely the 12 resistor and 6 resistor: () = + = 4 6 1 12 1 1 1 parallel
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Unformatted text preview: parallel R R Therefore, R eq = 14 + 4 = 18 . ______________________________________________________________________ 5 (3 marks) Given that the power absorbed by the 15 resistor is 15 W. Find the resistance R. Solution: Known quantities: Schematic; power absorbed by the 15 resistor. Find: Resistance of R. Analysis: The power dissipated by the 15- resistor is: P 15 = v 15 2 =15 W, therefore v 15 =15 V and i 1 = 1 A. Using the current dividing rule: A 1.5 ) (i 10 15 i 1 2 = = Applying KCL, we can find i R : i R = i 1 + i 2 = 2.5 A. Using KVL: -25 + 2.5R + 15 = 0, therefore, R = 4 ....
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This note was uploaded on 10/27/2009 for the course ECTE 290 taught by Professor Zhengli during the Three '09 term at University of Wollongong, Australia.

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Tutorial 2 - parallel R R Therefore, R eq = 14 + 4 = 18 .

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