Test+2--Fundamentals+of+Physics+II--key

Test+2--Fundamentals+of+Physics+II--key - Test...

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Unformatted text preview: Test 2—Fundamentals of Physics II --- Key 01. D 02. B 03. A 04. A 05. C 06. E 07. D 08. D 09. A 10. B 11. C 12. E 13. B 14. E 15. C 16. A 17. D 18. C 19. 20. Since we know that the current in the 8.00- resistor is 0.500 A, we can use Ohm's law (V Ω = IR) to find the voltage across the 8.00- resistor. The 8.00- resistor and the 16.0- resistor Ω Ω Ω are in parallel; therefore, the voltages across them are equal. Thus, we can also use Ohm's law to find the current through the 16.0-Ω resistor. The currents that flow through the 8.00-Ω and the 16.0-Ω resistors combine to give the total current that flows through the 20.0-Ω resistor. Similar reasoning can be used to find the current through the 9.00-Ω resistor. SOLUTION a. The voltage across the 8.00- resistor is V8=(0.500 A)(8.00 )=4.00 V=0.250A. Since Ω Ω this is also the voltage that is across the 16.0- resistor, we find that the current throughthis is also the voltage that is across the 16....
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This note was uploaded on 06/07/2009 for the course PHYSICS 2212 taught by Professor Geist during the Fall '09 term at Georgia Perimeter.

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Test+2--Fundamentals+of+Physics+II--key - Test...

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