# HW 15 - terry(ect328 homework 15 Turner(59130 This...

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terry (ect328) – homework 15 – Turner – (59130) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd 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. 247 V 2 k Ω 4 k Ω 494 V 8 k Ω 6 k a b c d e What is the current ±owing directly From “a” to “e”? Correct answer: 24 . 7 mA. Explanation: E 1 I 1 R 1 I 2 R 2 E 2 I ae I 4 R 4 3 R 3 a b c d e Let : R 1 = R = 2 k Ω , R 2 =2 R = 4 k Ω , R 3 =3 R = 6 k Ω , R 4 =4 R = 8 k Ω , E 1 = E = 247 V , and E 2 E = 494 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. Kirchho²’s Laws ± V = 0 around a closed loop . ± I = 0 at a circuit node . Solution: ³or this, we simply apply Kirchho²’s rules to the circuit. With the currents labeled as shown in the fgure above, we get the loop equations E - I 1 R - I 4 R 4 =0 2 I 2 R 2 - I 3 R 3 I 4 R 4 = I 3 R 3 , or I 1 R - I 4 (4 R ) = 0 (1) 2 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 ² E R - 4 I 4 ³ + ² E R - 2 I 4 ³ = ² 4 3 I 4 ³ + I 4 . Then I 4 = 6 25 E R ³inally, using the node equation at node “a”, we have I ae = I 4 - I 1 = 1 5 E R = 1 5 247 V 2 k Ω · 1 kΩ 1000 Ω · 1000 mA 1A = 24 . 7 mA .

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terry (ect328) – 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 Kirchho±’s 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 . When we insert this into equation (4) along with equation (6), we have V - I 1 R - ± I 1 + I 1 2 + V 2 R ² R 34 =0 . Solving for I 1 , I 1 = 1 25 V R . We then have I 2 = 13 25 V R I 34 = 14 25 V R . Finally V 34 = I 34 R 34 = 24 25 V. Returning to the original circuit (Fig. 2), we can obtain V c - V a = I 4 (4 R )= V 34 = 24 25 V = I 4 = 6 25 V R . Finally, examing node “a” as above, we have I ae = I 4 - I 1 = ± 6 25 - 1 25 ² V R = 1 5 V R = 24 . 7 mA . Or (at the node “e”) I ae = I 2 - I 3 = ± 13 25 - 8 25 ² V R = 1 5 V R = 24 . 7 mA . 002 10.0 points 3 Ω 4 Ω 6 Ω 8 Ω 11 Ω 1 Ω 2 Ω 13 V 23 V 37 V
terry (ect328) – homework 15 – Turner – (59130) 3 Find the magnitude of the current in the 13 V cell.

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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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HW 15 - terry(ect328 homework 15 Turner(59130 This...

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