_006 - Solution for Practice Test 6 for Test 2 Solution to...

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Unformatted text preview: Solution for Practice Test 6 for Test 2 Solution to Practice Test Problem 6.1(Reduce Resistance Network) Problem: For the circuit to the right, with R 1 = 100Ω , R 2 = 200Ω , and R 3 = 300Ω , and Δ V = 10 V . What is the current drawn by the circuit? Select One of the Following: (a-Answer) . 0667A (b) . 0167Ω (c) . 0183Ω (d) . 183Ω (e) . 033Ω a b Δ V R 1 R 3 R 2 Solution (a) Solve For Equivalent Resistance: The series combination of R 1 and R 2 can be replaced by its equivalent resistance, R s = R 1 + R 2 = 300Ω . The parallel combination R s and R 3 can then be reduced to give the equivalent resistance, 1 R ab = 1 R s + 1 R 3 = 1 300Ω + 1 300Ω giving R ab = 150Ω . (b) Compute Current: The current drawn by the circuit is given by Ohm’s Law, I ab = Δ V /R ab = 10 V/ 150Ω = . 0667 A . Total Points for Problem: 3 Points Solution to Practice Test Problem 6.2(Reduce Resistance Network) 1 Problem: For the circuit to the right, with R 1 = 100Ω , R 2 = 200Ω , and R 3 = 300Ω , and Δ V = 10 V . What is the voltage drop across R 1 ? Select One of the Following: (a) 6 . 67V (b) 10V (c) 1 . 67V (d-Answer) 3 . 33V (e) Can’t tell with information given. a b Δ V R 1 R 3 R 2 Solution (a) Compute Equivalent Resistance: The series combination of R 1 and R 2 can be replaced by its equivalent resistance, R s = R 1 + R 2 = 300Ω . The parallel combination R s and R 3 can then be reduced to give the equivalent resistance, 1 R ab = 1 R s + 1 R 3 = 1 300Ω + 1 300Ω giving R ab = 150Ω . (b) Compute Voltage Drop: The current through the series combination R 1 and R 2 is given by Ohm’s Law, I s = Δ V /R s = 10 V/ 300Ω = 0 . 0333 A . This current flows through both R 1 and R 2 because they are in series. The potential difference across R 1 is then Δ V 1 = I s R 1 = (0 . 0333 A )(100Ω) = 3 . 33 V . Total Points for Problem: 3 Points Solution to Practice Test Problem 6.3(Direction of Increase of Electric Potential) Problem: Examine the following list giving the relationship between the direction of field lines and the sign of potential difference. Select the answer which has the correct relation and best describes the physical reason for the relation. Select One of the Following: (a) The electric potential increases as the one moves at right angles to the field because this is the direction the field does most work on a positive charge. (b-Answer) The electric potential increases as one moves opposite the direction of the electric field because an external agent has to do work to move a positive charge in the direction opposite the field. (c) The electric potential is constant always because there really are minarets on the Hermitage. (d) Electric Potential Increases as one moves in the same direction as the electric field because the electric field does work on a positive charge as it moves along the field line....
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This note was uploaded on 05/04/2008 for the course PHYS 2074 taught by Professor Stewart during the Spring '08 term at Arkansas.

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_006 - Solution for Practice Test 6 for Test 2 Solution to...

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