sol-e2-sm09

sol-e2-sm09 - Phy 2049 Summer 2009 Exam II Solutions 1 A...

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Phy 2049 Summer 2009 Exam II Solutions 1. A circuit is drawn with four resistors of 4 Ohms and one resistor of 2 Ohm. An ideal 12V battery is connected to the circuit as shown. Two of the resistors are labeled A and B. What is the current through the resistor labeled B? The equivalent resistance is 6Ώ. Thus the current drawn from the battery is 12V/6Ώ = 2A. That is the current flowing through the resistor B. 2. In the figure in the problem above, what is the power dissipated in the resistor marked A? It is best done in plain words. There are 2A coming out of the battery. When they get to the junction, there is the same resistance on both branches. The current must split equally and therefore the current through the 2Ώ resistor is 1A. Finally the next junction also consists of equal resistances and the current splits equally. The current through the resistor A is ½A. The power absorbed I 2 R = 1W. The formal way to do this problem is to note: A) the current out of the battery must be the same as the current going into the battery which is same as the current through the resistor A. The potential difference across A is IR = 2x4 = 8V. B) Therefore the potential difference across the complex must be 4V. The resistance down the right arm is 4Ώ therefore the current through that arm is 1A. C) It goes through the 2Ώ resistor and is then split is two equal parts each fo ½A. .. The total power is IV = 2 x 12 = 24W. The power through resistor A is 16W. That through the 2Ώ resistor is 2W. Power in the 4Ώ on the left of the complex is 4W and in the two 4Ώ (parallel including resistor A) is 1W each. They all add up correctly. 3.

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This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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sol-e2-sm09 - Phy 2049 Summer 2009 Exam II Solutions 1 A...

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