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Unformatted text preview: benavides (jjb2356) homework 15 Turner (59130) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Note: Making use of the fact that the resistors and electric potentials are integer multiples of one another may make the solution less tedious. 235V 4 k 8 k 470V 1 6 k 1 2 k a b c d e What is the current flowing directly from a to e? Correct answer: 11 . 75 mA. Explanation: E 1 I 1 R 1 I 2 R 2 E 2 I ae I 4 R 4 I 3 R 3 a b c d e Let : R 1 = R = 4 k , R 2 = 2 R = 8 k , R 3 = 3 R = 12 k , R 4 = 4 R = 16 k , E 1 = E = 235 V , and E 2 = 2 E = 470 V . Note: Even though a and e are at the same electric potential (since they are connected by a wire), the wire can carry a current (since the wire has zero resistance). Basic Concepts: Resistors in Parallel and Series. Kirchhoffs Laws summationdisplay V = 0 around a closed loop . summationdisplay I = 0 at a circuit node . Solution: For this, we simply apply Kirchhoffs rules to the circuit. With the currents labeled as shown in the figure above, we get the loop equations E I 1 R I 4 R 4 = 0 2 E I 2 R 2 I 3 R 3 = 0 I 4 R 4 = I 3 R 3 , or E I 1 R I 4 (4 R ) = 0 (1) 2 E I 2 (2 R ) I 3 (3 R ) = 0 (2) I 4 (4 R ) = I 3 (3 R ) . (3) Solving these for the currents gives I 1 = E R 4 I 4 (1 ) I 2 = E R 3 2 I 3 (2 ) I 3 = 4 3 I 4 . (3 ) We can also immediately get I 2 = E R 2 I 4 . (2 ) Now we can use the node equation at node c I 1 + I 2 = I 3 + I 4 . Combining this with equations (1 ), (2 ), and (3 ) gives parenleftbigg E R 4 I 4 parenrightbigg + parenleftbigg E R 2 I 4 parenrightbigg = parenleftbigg 4 3 I 4 parenrightbigg + I 4 . Then I 4 = 6 25 E R Finally, using the node equation at node a, we have I ae = I 4 I 1 = 1 5 E R = 1 5 235 V 4 k 1 k 1000 1000 mA 1 A = 11 . 75 mA . benavides (jjb2356) homework 15 Turner (59130) 2 Solution: E 1 I 1 R 1 I 2 R 2 E 2 I 34 R 34 a b c d e Figure 3 We can reduce the number of loops and node equations if we simplify the circuit by noticing that the resistors 4 R and 3 R are parallel, as in Fig. 3. Then 1 R 34 = 1 R 3 + 1 R 4 = R 3 + R 4 R 3 R 4 R 34 = R 3 R 4 R 3 + R 4 = (3 R ) (4 R ) 3 R + 4 R = 12 7 R . Applying Kirchhoffs Laws to the two small loops and node c in this circuit, V I 1 R I 34 R 34 = 0 (4) 2 V I 2 (2 R ) I 34 R 34 = 0 (5) I 34 = I 1 + I 2 . (6) Subtracting equation (5) from equation (4) gives V I 1 R 2 V + I 2 (2 R ) = 0 . (7) Solving eqn. (7) for I 2 i gives I 2 = I 1 2 + V 2 R ....
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This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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